16. 3Sum Closest

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

 

int threeSumClosest(int* nums, int numsSize, int target) {
    int inner_target;
    int i = 0, j, k;
    int *solutionSet; //element in the returnArray
    int** returnArray = NULL;
    int diff;
    int min_diff = INT_MAX;
    int ret;
    
    quickSort(nums, 0, numsSize-1);
    
    for(; i < numsSize-2; i++){ //最外层遍历每个元素
        inner_target = target-nums[i];
        
        //里层: Two Sum
        j = i+1;
        k = numsSize-1;
        while(j<k){
            diff = abs(inner_target-nums[j]-nums[k]);
            if(diff < min_diff) {
                min_diff = diff;
                ret = nums[i]+nums[j]+nums[k];
            }
            
            if(nums[j]+nums[k] < inner_target) j++;
            else if(nums[j]+nums[k] > inner_target) k--;
            else return target;
        }
    }
    return ret;
}

void quickSort(int* nums, int start, int end){
    int p1 = start+1; 
    int p2 = end;
    int tmp;

    while(p1 <= p2){
        while(p1 <= p2 && nums[p1] <= nums[start]){
            p1++;
        }
        while(p1 <= p2 && nums[p2] > nums[start]){
            p2--;
        }
        if(p1 < p2){
            tmp = nums[p1];
            nums[p1] = nums[p2];
            nums[p2] = tmp;
            p1++;
            p2--;
        }
    }

    //put the sentinel at the end of the first subarray
    if(start!=p2){
    tmp = nums[start];
    nums[start] = nums[p2];
    nums[p2] = tmp;
    }
            
    if(start < p2-1) quickSort(nums,start, p2-1); //sort first subarray (<=sentinel)
    if(p1 < end) quickSort(nums,p1, end); //sort second subarray  (>sentinel)
}