[LeetCode] 16. 3Sum Closest 解题思路

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).

问题：给定一个数组和一个整数 n ，求数组中的三个元素，他们的和离 n 最近。

这道题和 3Sum 很相似，解题思路也很相似，先排序，然后采用双指针法依次比较。

由于 3Sum 很相似，思路就不详说。主要说下不同点：

1. 由于是求最接近值，当 s[i] + s[l] + s[r] - target == 0 的时候，即可返回结果。
2. 返回的结果是最接近 n 的三个元素之和。

int threeSumClosest(vector<int>& nums, int target) {
    
    std::sort(nums.begin(), nums.end());
    
    int closest = target - nums[0] - nums[1] - nums[2];
    
    for (int i = 0 ; i < nums.size(); i++) {
        
        int l = i + 1;
        int r = (int)nums.size() - 1;
        
        int newTarget = target - nums[i];
        
        while (l < r) {
            if (nums[l] + nums[r] == newTarget) {
                return target;
            }
            
            int diff = newTarget - nums[l] - nums[r];
            if (abs(diff) < abs(closest)) {
                closest = diff;
            }
            
            if (nums[l] + nums[r] < newTarget){
                l++;
            }else{
                r--;
            }
        }
    }
    
    return target - closest;
}