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@Yano 2017-12-10T01:57:29.000000Z 字数 1506 阅读 926

LeetCode 473 Matchsticks to Square

LeetCode


题目描述

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

Example 2:

Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.

Note:

The length sum of the given matchsticks is in the range of 0 to 10^9.
The length of the given matchstick array will not exceed 15.

题目分析

题目的问题抽象出来就是:在n个数中能否找出4组数,使得他们的和相等。

题目显然是一个递归问题,循环条件是4组数的和,返回条件是将n个数都用到后,每组数的和是否相等。

注意

  1. 应该先对数组排序,并从大到小进行递归,因为两个大数相加更可能超过正方形的边长,所以递归深度更少。否则会超时。
  2. 在第一次做时,我额外用了boolean数组来判断nums的对应元素是否使用过,但是这些操作在递归中就包括了。

代码

  1. public boolean makesquare(int[] nums) {
  2. if (nums == null || nums.length < 4) {
  3. return false;
  4. }
  5. int tmp = 0;
  6. for (int i : nums) {
  7. tmp += i;
  8. }
  9. if (tmp % 4 != 0) {
  10. return false;
  11. }
  12. Arrays.sort(nums);
  13. return robot(0, nums, new int[4], tmp / 4);
  14. }
  15. private boolean robot(int pos, int[] nums, int[] sum, int each) {
  16. if (pos == nums.length) {
  17. return sum[0] == each && sum[1] == each && sum[2] == each;
  18. }
  19. for (int i = 0; i < 4; i++) {
  20. int index = nums.length - 1 - pos;
  21. if (sum[i] + nums[index] > each) {
  22. continue;
  23. }
  24. sum[i] += nums[index];
  25. if (robot(pos + 1, nums, sum, each)) {
  26. return true;
  27. }
  28. sum[i] -= nums[index];
  29. }
  30. return false;
  31. }
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