为什么80%的码农都做不了架构师？>>> 
问题：
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
0to10 ^ 9. - The length of the given matchstick array will not exceed
15.
解决：
① 跟之前有道题Partition Equal Subset Sum有点像，那道题问我们能不能将一个数组分成和相等的两个子数组，而这道题实际上是让我们将一个数组分成四个和相等的子数组。
先给数组从大到小的顺序排序，这样大的数字先加，如果超过target了，就直接跳过了后面的再次调用递归的操作。
我们建立一个长度为4的数组sums来保存每个边的长度和，我们希望每条边都等于target，数组总和的四分之一。然后我们遍历sums中的每条边，我们判断如果加上数组中的当前数字大于target，那么我们跳过，如果没有，我们就加上这个数字，然后对数组中下一个位置调用递归，如果返回为真，我们返回true，否则我们再从sums中对应位置将这个数字减去继续循环。
class Solution { //39ms
public boolean makesquare(int[] nums) {
if (nums == null || nums.length < 4) return false;
int sum = 0;
for (int n : nums){
sum += n;
}
if (sum % 4 != 0) return false;
int[] sums = new int[4];
Arrays.sort(nums);
return dfs(nums,sums,nums.length - 1,sum / 4);
}
public boolean dfs(int[] nums,int[] sums,int i,int target){
if (i < 0){
return sums[0] == target && sums[1] == target && sums[2] == target;
}
for (int j = 0;j < 4;j ++){
if (sums[j] + nums[i] > target) continue;
sums[j] += nums[i];
if (dfs(nums,sums,i - 1,target)) return true;
sums[j] -= nums[i];
}
return false;
}
}