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).

跟 3sum smallest类似，O(n^2)的复杂度。第一个数从0开始向后遍历，在这期间第二个数初始化为i + 1，第三个数初始化为length - 1。如果三个数的和大于target，high--，如果小于，low++。比较现在的和跟之前的和是不是更接近target，更接近的话替换之前的和。代码如下：

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int min = Integer.MAX_VALUE;
        int res = 0;
        for (int i = 0; i < nums.length; i ++) {
            int low = i + 1, high = nums.length - 1, sum = target - nums[i];
            while (low < high) {
                if (min > Math.abs(nums[low] + nums[high] - sum)) {
                    res = nums[low] + nums[high] + nums[i];
                    min = Math.abs(nums[low] + nums[high] - sum);
                }
                if (nums[low] + nums[high] == sum) {
                    return target;
                } else if (nums[low] + nums[high] < sum) {
                    low ++;
                } else {
                    high --;
                    while (low + 1 < high && nums[high] == nums[high + 1]) high --;
                }
            }
        }
        return res;
    }
}

上面算法用的是差来比较是不是更接近，也可以用存储和的形式来判断是不是更接近，代码如下：

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        int result = num[0] + num[1] + num[num.length - 1];
        Arrays.sort(num);
        for (int i = 0; i < num.length - 2; i++) {
            int start = i + 1, end = num.length - 1;
            while (start < end) {
                int sum = num[i] + num[start] + num[end];
                if (sum > target) {
                    end--;
                } else {
                    start++;
                }
                if (Math.abs(sum - target) < Math.abs(result - target)) {
                    result = sum;
                }
            }
        }
        return result;
    }
}