@Yano 2015-12-30T11:28:03.000000Z 字数 10468 阅读 9551

LeetCode

LeetCode 题目汇总

# Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its level order traversal as:

[  [3],  [9,20],  [15,7]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

1. 通过统计每一行的结点数

toBePrinted：当前待打印结点的数量
nextLevel：下一层的结点数量

2. 插入特殊结点

a queue stores [step0, step1, step2, ...]

while queue is not empty

current_step = queue.poll()

// do something here with current_step
// like counting



    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> rt = new ArrayList<List<Integer>>();        if (root == null) {            return rt;        }        Deque<TreeNode> deque = new LinkedList<TreeNode>();        deque.add(root);        int toBePrinted = 1;        int nextLevel = 0;        List<Integer> level = new LinkedList<Integer>();        while (!deque.isEmpty()) {            TreeNode p = deque.poll();            level.add(p.val);            toBePrinted--;            if (p.left != null) {                deque.addLast(p.left);                nextLevel++;            }            if (p.right != null) {                deque.addLast(p.right);                nextLevel++;            }            if (toBePrinted == 0) {                toBePrinted = nextLevel;                nextLevel = 0;                rt.add(new ArrayList<Integer>(level));                level.clear();            }        }        return rt;    }

    public List<List<Integer>> levelOrder2(TreeNode root) {        List<List<Integer>> rt = new ArrayList<List<Integer>>();        if (root == null) {            return rt;        }        final TreeNode END = new TreeNode(0);        Deque<TreeNode> deque = new LinkedList<TreeNode>();        List<Integer> level = new LinkedList<Integer>();        deque.add(root);        deque.add(END);        while (!deque.isEmpty()) {            TreeNode p = deque.pop();            if (p == END) {                rt.add(new ArrayList<Integer>(level));                level.clear();                if (!deque.isEmpty()) {                    deque.add(END);                }            } else {                level.add(p.val);                if (p.left != null) {                    deque.add(p.left);                }                if (p.right != null) {                    deque.add(p.right);                }            }        }        return rt;    }

# Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Collections.reverse(result);
    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> rt = new ArrayList<List<Integer>>();        if (root == null) {            return rt;        }        final TreeNode END = new TreeNode(0);        Deque<TreeNode> deque = new LinkedList<TreeNode>();        List<Integer> level = new LinkedList<Integer>();        deque.add(root);        deque.add(END);        while (!deque.isEmpty()) {            TreeNode p = deque.pop();            if (p == END) {                rt.add(new ArrayList<Integer>(level));                level.clear();                if (!deque.isEmpty()) {                    deque.add(END);                }            } else {                level.add(p.val);                if (p.left != null) {                    deque.add(p.left);                }                if (p.right != null) {                    deque.add(p.right);                }            }        }        Collections.reverse(rt);        return rt;    }

# Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> rt = new ArrayList<List<Integer>>();        if (root == null) {            return rt;        }        final TreeNode END = new TreeNode(0);        Deque<TreeNode> deque = new LinkedList<TreeNode>();        List<Integer> level = new LinkedList<Integer>();        int count = 0;        deque.add(root);        deque.add(END);        while (!deque.isEmpty()) {            TreeNode p = deque.pop();            if (p == END) {                if (count % 2 == 1) {                    Collections.reverse(level);                }                count++;                rt.add(new ArrayList<Integer>(level));                level.clear();                if (!deque.isEmpty()) {                    deque.add(END);                }            } else {                level.add(p.val);                if (p.left != null) {                    deque.add(p.left);                }                if (p.right != null) {                    deque.add(p.right);                }            }        }        return rt;    }

# Course Schedule

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

BFS:

    // BFS    public static boolean canFinish(int numCourses, int[][] prerequisites) {        // 参数检查        if (prerequisites == null) {            return false;        }        int len = prerequisites.length;        if (numCourses <= 0 || len == 0) {            return true;        }        // 记录每个course的prerequisites的数量        int[] pCounter = new int[numCourses];        for (int i = 0; i < len; i++) {            pCounter[prerequisites[i][0]]++;        }        // 用队列记录可以直接访问的course        LinkedList<Integer> queue = new LinkedList<Integer>();        for (int i = 0; i < numCourses; i++) {            if (pCounter[i] == 0) {                queue.add(i);            }        }        // 取出队列的course，判断        int numNoPre = queue.size();        while (!queue.isEmpty()) {            int top = queue.remove();            for (int i = 0; i < len; i++) {                // 该course是某个course的prerequisites                if (prerequisites[i][1] == top) {                    pCounter[prerequisites[i][0]]--;                    if (pCounter[prerequisites[i][0]] == 0) {                        numNoPre++;                        queue.add(prerequisites[i][0]);                    }                }            }        }        return numNoPre == numCourses;    }

DFS:

    // DFS    public static boolean canFinish2(int numCourses, int[][] prerequisites) {        // 参数检查        if (prerequisites == null) {            return false;        }        int len = prerequisites.length;        if (numCourses <= 0 || len == 0) {            return true;        }        int[] visit = new int[numCourses];        // key：course；value：以该course为prerequisites的course        HashMap<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();        // 初始化map        for (int[] p : prerequisites) {            if (map.containsKey(p[1])) {                map.get(p[1]).add(p[0]);            } else {                ArrayList<Integer> l = new ArrayList<Integer>();                l.add(p[0]);                map.put(p[1], l);            }        }        // dfs        for (int i = 0; i < numCourses; i++) {            if (!canFinishDFS(map, visit, i)) {                return false;            }        }        return true;    }    private static boolean canFinishDFS(            HashMap<Integer, ArrayList<Integer>> map, int[] visit, int i) {        if (visit[i] == -1) {            return false;        }        if (visit[i] == 1) {            return true;        }        visit[i] = -1;        // course i是某些course的prerequisites        if (map.containsKey(i)) {            for (int j : map.get(i)) {                if (!canFinishDFS(map, visit, j)) {                    return false;                }            }        }        visit[i] = 1;        return true;    }

# Course Schedule II

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

Hints:

1. This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

    public int[] findOrder(int numCourses, int[][] prerequisites) {        // 参数检查        if (prerequisites == null) {            throw new IllegalArgumentException();        }        int len = prerequisites.length;        if (len == 0) {            int[] seq = new int[numCourses];            for (int i = 0; i < seq.length; i++) {                seq[i] = i;            }            return seq;        }        int[] seq = new int[numCourses];        int c = 0;        // 记录每个course的prerequisites的数量        int[] pCounter = new int[numCourses];        for (int i = 0; i < len; i++) {            pCounter[prerequisites[i][0]]++;        }        // 用队列记录可以直接访问的course        LinkedList<Integer> queue = new LinkedList<Integer>();        for (int i = 0; i < numCourses; i++) {            if (pCounter[i] == 0) {                queue.add(i);            }        }        // 取出队列的course，判断        int numNoPre = queue.size();        while (!queue.isEmpty()) {            int top = queue.remove();            // 保存结果 +_+            seq[c++] = top;            for (int i = 0; i < len; i++) {                // 该course是某个course的prerequisites                if (prerequisites[i][1] == top) {                    pCounter[prerequisites[i][0]]--;                    if (pCounter[prerequisites[i][0]] == 0) {                        numNoPre++;                        queue.add(prerequisites[i][0]);                    }                }            }        }        if (numNoPre != numCourses) {            return new int[] {};        }        return seq;    }

# Surrounded Regions

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

X X X XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

    static class Point {        int x;        int y;        Point(int x, int y) {            this.x = x;            this.y = y;        }    }    HashSet<String> boarderConnected;    String pointId(int x, int y) {        return x + "," + y;    }    boolean connectIfNotConnected(char[][] board, int x, int y) {        if (x < 0 || y < 0)            return false;        if (x >= board.length || y >= board[0].length)            return false;        if (board[x][y] == 'X')            return false;        String id = pointId(x, y);        if (boarderConnected.contains(id))            return false;        boarderConnected.add(id);        return true;    }    void connectBoarder(char[][] board, int x, int y) {        LinkedList<Point> queue = new LinkedList<Point>();        queue.add(new Point(x, y));        while (!queue.isEmpty()) {            Point p = queue.poll();            if (connectIfNotConnected(board, p.x, p.y)) {                queue.add(new Point(p.x + 1, p.y));                queue.add(new Point(p.x - 1, p.y));                queue.add(new Point(p.x, p.y + 1));                queue.add(new Point(p.x, p.y - 1));            }        }    }    public void solve(char[][] board) {        int mx = board.length;        if (mx < 3)            return;        int my = board[0].length;        if (my < 3)            return;        boarderConnected = new HashSet<String>();        int x;        int y;        for (x = 0; x < mx; x++) {            connectBoarder(board, x, 0);            connectBoarder(board, x, my - 1);        }        for (y = 0; y < my; y++) {            connectBoarder(board, 0, y);            connectBoarder(board, mx - 1, y);        }        for (x = 0; x < mx; x++) {            for (y = 0; y < my; y++) {                if (board[x][y] == 'O') {                    if (!boarderConnected.contains(pointId(x, y))) {                        board[x][y] = 'X';                    }                }            }        }    }

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