LeetCode_3 sum closet

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).

3 sum 的变形,这里不需要考虑重复而已

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(), num.end());
        int len = num.size();
        if(len < 3) return 0;
        int res = num[0] + num[1] + num[2];
        for(int i = 0; i <= len -3; ++i){
            
            int low = i+1;
            int high = len -1;
            while(low < high){
                
                int sum = num[i] + num[low] + num[high];
                if(sum == target)
                    return target;
                else if(sum < target){
                    ++low;
                    if(abs(sum - target) < abs(res - target))
                        res = sum;
                }else{
                    --high;
                    if(abs(sum - target) < abs(res - target))
                        res = sum;

                }
            }
        }
        
        return res;
    }
};
原文地址:https://www.cnblogs.com/graph/p/3343847.html