力扣-----三數和爲0

鏈接：https://leetcode-cn.com/problems/3sum

 

給你一個包含 n 個整數的數組 nums，判斷 nums 中是否存在三個元素 a，b，c ，使得 a + b + c = 0 ？請你找出所有滿足條件且不重複的三元組。

注意：答案中不可以包含重複的三元組。

 

示例：

給定數組 nums = [-1, 0, 1, 2, -1, -4]，

滿足要求的三元組集合爲：
[
  [-1, 0, 1],
  [-1, -1, 2]
]

 

int CompareByIncrease(const void* a, const void* b)
{
    return *(int*)a - *(int*)b; // 快排構造遞增序列
}

int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){

    *returnSize = 0;

    if( numsSize < 3 )
    {
        return NULL;
    }
    
    int cur = 0;
    int low = cur + 1;
    int high = numsSize - 1;
    

    int** returnArray = (int**)malloc(sizeof(int*) * (numsSize)*(numsSize));
    *returnColumnSizes = (int*)malloc(sizeof(int) * (numsSize)*(numsSize)); 

    qsort(nums, numsSize, sizeof(int), CompareByIncrease);
    

    while((nums[cur] <= 0) && (cur + 1 < numsSize - 1)) 
    {
 
        low = cur + 1;
        high = numsSize - 1;
        
        while(low < high)
        {
            if(0 == (nums[low] + nums[high] + nums[cur]))
            {
                returnArray[*returnSize] = (int*)malloc(sizeof(int) * 3); // 每次找到一組，二級指針就分配三個空間
                (*returnColumnSizes)[*returnSize] = 3; // 記錄列數，非常騷而弔詭的操作
                returnArray[*returnSize][0] = nums[cur];
                returnArray[*returnSize][1] = nums[low];
                returnArray[*returnSize][2] = nums[high];
                (*returnSize)++;
                
                // 去low和high的重,非常不規範的寫法，但看着極端舒服
                while( (nums[low] == nums[++low]) && (low < high) ); // 往後去重
                while( (nums[high] == nums[--high]) && (low < high) ); // 往前去重
            }
            else if(0 < (nums[low] + nums[high] + nums[cur]))
            {
                high--;
            }
            else
            {
                low++;
            }
        }
    
        while( (nums[cur] == nums[++cur]) && (cur + 1 < numsSize - 1) );  
    }
    return returnArray;
}