15. 三数之和
15. 三数之和
难度中等
给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？请你找出所有和为 0 且不重复的三元组。
注意：答案中不可以包含重复的三元组。
示例 1：
输入：nums = [-1,0,1,2,-1,-4] 输出：[[-1,-1,2],[-1,0,1]]
示例 2：
输入：nums = [] 输出：[]
示例 3：
输入：nums = [0] 输出：[]
提示：
0 <= nums.length <= 3000-105 <= nums[i] <= 105
我的解法：暴力，但会超时 315/318,
就这破算法也写了3个小时，丑的不忍直视。去重耗时间
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int cmp(const void*a, const void*b){
return *(int*)a - *(int*)b;
}
int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
int i = 0;
int j = 0;
int k = 0;
int m = 0;
int sum = 0;
int arr[3000][3] = {0};
int cnt = 0;
int tmp = 0;
int** res = NULL;
if(!nums){
return res;
}
qsort(nums, numsSize,sizeof(nums[0]),cmp);//排序
for(i=0; i<numsSize; i++){
for(j=i+1; j<numsSize; j++){
for(k = j+1;k<numsSize;k++){
tmp = nums[i]+nums[j]+nums[k];
if(tmp == 0){
arr[cnt][0] = nums[i];
arr[cnt][1] = nums[j];
arr[cnt][2] = nums[k];
cnt++;
for(m =0;m<cnt-1;m++){ //for循环去重，用最新添加的一组和前面的cnt -2组对比，
if(arr[m][0]==nums[i]&& arr[m][1] == nums[j]){ //对比前两个数，如果相同，就将cnt-- ，最新添加的那组不算
cnt--;
}
}
}
}
}
}
*returnSize = cnt;
*returnColumnSizes = malloc(cnt*sizeof(int));
res = (int**)malloc(cnt*sizeof(int*));
for(i=0;i<cnt;i++){
(*returnColumnSizes)[i] = 3;
res[i] = malloc(3*sizeof(int));
res[i][0] = arr[i][0];
res[i][1] = arr[i][1];
res[i][2] = arr[i][2];
}
return res;
}
官方解法: 双指针
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int cmp(const void*a, const void*b){
return *(int*)a - *(int*)b;
}
int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
int i = 0;
int l = 1;
int r = numsSize -1;
int m = 0;
int sum = 0;
int arr[20000][3] = {0};
int cnt = 0;
int target = 0;
int** res = NULL;
if(!nums){
return res;
}
qsort(nums, numsSize,sizeof(nums[0]),cmp);
for(i=0; i<numsSize; i++){
if(i>0 && nums[i] == nums[i-1]){
continue;
}
target = 0-nums[i];
l = i+1;
r = numsSize-1;
while(l<r){
sum = nums[l]+nums[r];
if(sum == target){
arr[cnt][0] = nums[i];
arr[cnt][1] = nums[l];
arr[cnt][2] = nums[r];
cnt++;
while(l<r && nums[l]==nums[l+1]){
l++;
}
l++;
while(l<r && nums[r]==nums[r-1]){
r--;
}
r--;
}else if(sum>target){
r--;
}else {
l++;
}
}
}
*returnSize = cnt;
*returnColumnSizes = malloc(cnt*sizeof(int));
res = (int**)malloc(cnt*sizeof(int*));
for(i=0;i<cnt;i++){
(*returnColumnSizes)[i] = 3;
res[i] = malloc(3*sizeof(int));
res[i][0] = arr[i][0];
res[i][1] = arr[i][1];
res[i][2] = arr[i][2];
}
return res;
}
