【LeetCode】LeetCode——第16题：3Sum Closest

16. 3Sum Closest

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Total Accepted: 76085  Total Submissions: 261853  Difficulty: Medium

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).

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题目的大概意思是：给定一个数组nums和一个整数target，从数组中找出三个数，使它们的和最接近target，并返回这个和值。

这道题难度等级：中等

思路：

1、与题目3Sum类似，也是先将数组进行排序，然后枚举和夹逼，用minSum来记录最接近target的值；

2、从左往右枚举，枚举位置为i(0，左、右分别用l、r来表示，且l，r>i，从两边往中间夹逼(这里的中间指的是枚举位置)；

3、tmp=nums[l]+nums[i]+nums[r]，若abs(tmp-target)<(minSum-target)，说明找到个更接近的，要记录下来。并且，若tmp>target，则说明和值偏大，此时--r；若tmp，说明和值偏小，则++l；若tmp==target，说明找到个无偏差的，直接return了。

代码如下：

class Solution {
public:
    int threeSumClosest(vector& nums, int target) {
        sort(nums.begin(), nums.end());
		int minSum = 99999999;
		int l, r, tmp;
		for (int i = 1; i < nums.size() - 1; ++i){
			l = 0;r = nums.size() - 1;
			while(l < i && r > i){
				tmp = nums[l] + nums[i] + nums[r];
				if (tmp == target){return target;}
				(tmp > target) ? (--r) : (++l);
				if (abs(minSum - target) > abs(tmp - target)){
					minSum = tmp;
				}
			}
		}
		return minSum;
    }
};

提交代码后，顺利AC掉该题，Runtime: 12 ms。

还有另外一种能AC该题的办法，当然跟上面的差不多，只不过在夹逼的时候，是从中间往两边发散。

代码如下：

class Solution {
public:
	int threeSumClosest(vector& nums, int target) {
		sort(nums.begin(), nums.end());
		int minSum = 99999999;
		int l, r, tmp;
		for (int i = 1; i < nums.size() - 1; ++i){
			l = i - 1;
			r = i + 1;
			while(l >= 0 && r < nums.size()){
				tmp = nums[l] + nums[i] + nums[r];
				if (tmp == target){return target;}
				(tmp > target) ? (--l): ++r;
				if (abs(tmp - target) < abs(minSum - target)){
					minSum = tmp;
				}
			}
		}
		return minSum;
	}
};

Runtime: 16 ms