【题目】

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).

【解析】

和 3Sum解题报告 非常像，与之不同的是，不再是求三个数的和是不是为0，而是看三个数的和与target的差是否为最小，仅仅需记录当前最优解并不断更新其值就可。

【Java代码】O(n^2)

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        if (num == null || num.length < 3) return 0;
        
        Arrays.sort(num);
        
        int ret = 0;
        int closestDist = Integer.MAX_VALUE;
        int len =  num.length;
        for (int i = 0; i < len-2; i++) {
            if (i > 0 && num[i] == num[i-1]) continue;
            
            int l = i+1, r = len-1;
            while (l < r) {
                int sum = num[i] + num[l] + num[r];
                if (sum < target) {
                    if (target-sum < closestDist) {
                        closestDist = target - sum;
                        ret = sum;
                    }
                    l++;
                } else if (sum > target) {
                    if (sum-target < closestDist) {
                        closestDist = sum - target;
                        ret = sum;
                    }
                    r--;
                } else { //when sum == target, return sum.
                    return sum;
                }
            }
        }
        
        return ret;
    }
}

easy出错的地方是。把 ret 初始值设为 Integer.MAX_VALUE。然后后面计算 closestDist = Math.abs(ret - target)，这样会导致溢出！。