#15 3Sum

题目链接：https://leetcode.com/problems/3sum/

Given an array S of n integers, are there elements a, b, c in S such that a + b + c =
0? Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:
(-1, 0, 1)
(-1, -1, 2)

/**
* Return an array of arrays of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
void Sort(int *data,int n) {<span style="white-space:pre">			</span>//归并排序非递归方式实现
int *tmp = (int *)malloc(n * sizeof(int));
int lBegin,lEnd,rBegin,rEnd;
int i,j,k;
int len = 1;
while (len < n) {
lBegin = 0;
lEnd = len - 1;
rBegin = len;
while (rBegin < n) {
rEnd = lEnd + len < n - 1 ? lEnd + len : n - 1;
i = lBegin,j = rBegin,k = lBegin;
while (i <= lEnd && j <= rEnd) {
if (data[i] <= data[j])
tmp[k++] = data[i++];
else
tmp[k++] = data[j++];
}
while (i <= lEnd)
tmp[k++] = data[i++];
while (j <= rEnd)
tmp[k++] = data[j++];
for (i = lBegin; i <= rEnd; ++i)
data[i] = tmp[i];
lBegin += 2 * len;
lEnd += 2 * len;
rBegin += 2 * len;
}
len *= 2;
}
free(tmp);
}

int** threeSum(int* nums,int numsSize,int* returnSize) {
int i,j,k;
int **ret = (int **)malloc(sizeof(int *) * 200);
for (i = 0; i < 200; ++i)
ret[i] = (int *)malloc(sizeof(int) * 3);
*returnSize = 0;
Sort(nums,numsSize);<span style="white-space:pre">		</span>//排序
for (i = 0; i < numsSize; ++i) {<span style="white-space:pre">		</span>//每趟固定第一个数字，另外两个数分别从生下数列的头和尾向中间逼近
if(i > 0 && nums[i] == nums[i - 1])<span style="white-space:pre">		</span>//第一个数相等，之后所求解会与上一趟所求解重合，直接跳过
continue;
j = i + 1;<span style="white-space:pre">				</span>//第二个数从小变大
k = numsSize - 1;<span style="white-space:pre">			</span>//第三个数从大变小
while (j < k) {
if (nums[i] + nums[j] + nums[k] < 0)<span style="white-space:pre">	</span>//和小于0，需要增大sum，即第二个数向右移动从而增大
++j;
else if (nums[i] + nums[j] + nums[k] > 0)<span style="white-space:pre">	</span>//和大于0，第三个数向左移动
--k;
else {
if (*returnSize == 0 || ret[*returnSize - 1][0] != nums[i] || ret[*returnSize - 1][1] != nums[j] || ret[*returnSize - 1][2] != nums[k]) {<span style="white-space:pre">				</span>//取出重复解。 因为数组排过序，重复解只可能与上一个解相同，只需要比较上一次所求解
ret[*returnSize][0] = nums[i];
ret[*returnSize][1] = nums[j];
ret[*returnSize][2] = nums[k];
++*returnSize;
}
++j;
--k;
}
}
}
return ret;
}