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).

https://leetcode.com/problems/3sum-closest/#/solutions

http://www.cnblogs.com/EdwardLiu/p/4012459.html

先将数组排个序，然后开始遍历数组，思路跟那道三数之和很相似，都是先确定一个数，然后用两个指针left和right来滑动寻找另外两个数，每确定两个数，我们求出此三数之和，然后算和给定值的差的绝对值存在newDiff中，然后和diff比较并更新diff和结果closest即可:

这道题跟3Sum很像，区别就是要维护一个最小的diff，求出和目标最近的三个和。brute force时间复杂度为O(n^3)，优化的解法是使用排序之后2 pointers的方法:

通过排序后的和与零比较, 两头的指针移动少计算不符合的项.

总的时间复杂度为O(n^2+nlogn)=(n^2),空间复杂度是O(n)

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        if (num == null || num.length < 3) return 0;
        Arrays.sort(num);
        int minDiff = num[0] + num[1] + num[2] - target;
        int diff = 0;
        for (int i=num.length-1; i>=2; i--) {
            if (i<num.length-1 && num[i]==num[i+1]) continue;
            diff = twoSumClosest(num, 0, i-1, target-num[i]);
            if (Math.abs(diff) < Math.abs(minDiff)) {
                minDiff = diff;
            }
        }
        return minDiff + target;
    }
    
    public int twoSumClosest(int[] num, int l, int r, int tar) {
        int minDif = num[l] + num[r] - tar;
        int dif = 0;
        while (l < r) {
            dif = num[l] + num[r] - tar;
            if (dif == 0) return dif;
            if (Math.abs(dif) < Math.abs(minDif)) {
                minDif = dif;
            }
            if (dif < 0) {
                l++;
            }
            else {
                r--;
            }
        }
        return minDif;
    }
}

　　for (int i = 0; i < nums.length - 2; i++) {

l++;

r--;

l < r