3Sum Closest : https://leetcode.com/problems/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).
和 上篇 3Sum 思想相同,利用夹逼思想,枚举一个数nums[i],从j=i+1, k = nums.size()-1;夹逼,sum小于target:j++,sum大于target,k–
此问题不需考虑重复,因为只需要一个和的结果。
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int result = nums[0]+nums[1]+nums[2];//初始化
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size()-2; i++) {
int j = i+1;
int k = nums.size()-1;
int sum;
while (j < k) {
sum = nums[i]+nums[j]+nums[k];
if (abs(sum-target) < abs(result-target))
result = sum;
if (sum < target)
j++; //不需检测nums[j]==nums[j-1],可以重复,不需保存满足条件的元素
else if (sum > target)
k--;
else
return target;
}
//注:不可将第12、13行放在此处,因为此处的终止条件是j == k
}
return result;
}
};
原文地址:http://blog.csdn.net/quzhongxin/article/details/46117629