Leetcode:3Sum Closest

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).

先对给定的数组排序（升序），然后依次遍历每个数组中的每个元素，并另外定义两个指针，分别指向遍历到数的后一个元素，和数组的最后一个元素。计算这三个数字之和，如果比目标数值大，则右边的指针左移一位；如果比目标数值小，则左边的指针右移一位。遍历的同时记录最接近目标数值的和，最后返回该值。如果遍历过程中遇到等于目标数值的情况，则直接返回。

实现代码：

class Solution {public:    int threeSumClosest(vector<int> &num, int target) {    vector<int> v(num.begin(), num.end()); // I didn't wanted to disturb original array.    int n = 0;    int ans = 0;    int sum;    sort(v.begin(), v.end());    // If less then 3 elements then return their sum    while (v.size() <= 3) {        return accumulate(v.begin(), v.end(), 0);    }    n = v.size();    /* v[0] v[1] v[2] ... v[i] .... v[j] ... v[k] ... v[n-2] v[n-1]     *                    v[i]  <=  v[j]  <= v[k] always, because we sorted our array.      * Now, for each number, v[i] : we look for pairs v[j] & v[k] such that      * absolute value of (target - (v[i] + v[j] + v[k]) is minimised.     * if the sum of the triplet is greater then the target it implies     * we need to reduce our sum, so we do K = K - 1, that is we reduce     * our sum by taking a smaller number.     * Simillarly if sum of the triplet is less then the target then we     * increase out sum by taking a larger number, i.e. J = J + 1.     */    ans = v[0] + v[1] + v[2];    for (int i = 0; i < n-2; i++) {        int j = i + 1;        int k = n - 1;        while (j < k) {            sum = v[i] + v[j] + v[k];            if (abs(target - ans) > abs(target - sum)) {                ans = sum;                if (ans == target) return ans;            }            (sum > target) ? k-- : j++;        }    }    return ans;}};