【LeetCode】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).

java code :  复杂度 O(n^2 logn) 最内层用二分查找

public class Solution {    public int threeSumClosest(int[] num, int target) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if(num.length < 3)return 0;Arrays.sort(num);int sum = num[0] + num[1] + num[2];int dist = Math.abs((num[0] + num[1] + num[2] - target));if(dist == 0)return sum;for(int i = 0; i < num.length - 2; i++){for(int j = i + 1; j < num.length -1; j++){int val = target - num[i] - num[j];int index = binarysearch(num, val, j+1, num.length - 1);if(num[index] == val)return target;else {int sum1 = num[i] + num[j] + num[index];int dist1 = Math.abs((target - sum1));if(dist1 < dist){sum = sum1;dist = dist1;}if(index < num.length -1){int sum2 = num[i] + num[j] + num[index + 1];int dist2 = Math.abs((target - sum2));if(dist2 < dist){sum = sum2;dist = dist2;}}}}}return sum;    }    public int binarysearch(int[] array, int key, int st, int end){int lhs = st, rhs = end;if(st == end)return st;while(lhs <= rhs){int mid = (lhs + rhs) >> 1;if(array[mid] == key)return mid;else if(array[mid] < key)lhs = mid + 1;else rhs = mid - 1;}if(rhs < st)return st;return rhs;}}