LeetCode 16: 3Sum Closest

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).

解题思路

和3Sum的思路类似。先对数组排序；枚举第一个数的位置i，然后设两个指针分别从数组的 i+1 和 numsLen-1 位置开始向中间收缩，if sum > target则右指针往左移， if sum < target则左指针往右移，在此过程中需要更新结果。时间复杂度为$O(N^2)$，代码如下：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end()); // 对数组排序

        int numsLen = nums.size();
        int result, minDiff = INT_MAX;

        for (int i = 0; i < numsLen; ++i) {
            // 先固定第一个数
            int t = target - nums[i];

            // 设两个指针，分别从数组的 i+1 和 numsLen-1 位置开始向中间收缩
            int left = i + 1;
            int right = numsLen - 1;
            while (left < right) {
                int sum2 = nums[left] + nums[right];
                if (abs(sum2 - t) < minDiff) {
                    // 更新结果
                    minDiff = abs(sum2 - t);
                    result = nums[i] + sum2;
                }

                // 判断移动 left 还是 right 指针
                if (sum2 < t) {
                    left++;
                }
                else {
                    right--;
                }
            }
        }
        return result;
    }
};