LeetCode[16]： 3Sum Closest

题目描述

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).

解题思路

这道题和之前的两数之和的解决方式类似，区别是在其基础上增加了第三个数。我们可以通过固定其中一个数numsi$num{s}_i$，然后寻找两数之和最接近target−numsi$target-num{s}_i$，将问题转化为两数之和的问题。首先将数据排序，遍历数组固定numsi$num{s}_i$然后使j=i+1，k=length−1$j=i+1，k=length-1$，通过移动j，k$j，k$两个下标，来使numsj+numsk逼近target−numsi$num{s}_j+num{s}_k逼近target-num{s}_i$。然后保存当前最优解和当前三数之和更接近目标target的值。

代码

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int threeSum = nums[0] + nums[1] + nums[2];
        int closestNum = threeSum;
        //先排序
        Arrays.sort(nums);
        for (int i=0; i < nums.length-2; i++) {
            int twoSum = target - nums[i];
            int j = i + 1, k = nums.length-1;
            while (j < k) {
                threeSum = nums[i] + nums[j] + nums[k];
                if(Math.abs(threeSum-target) < Math.abs(closestNum-target))
                    closestNum = threeSum;
                if (nums[j] + nums[k] < twoSum) {
                    j++;
                } else if (nums[j] + nums[k] > twoSum) {
                    k--;
                } else { return target; }
            }
        }
        return closestNum;
    }
}