16. 3Sum Closest -Medium

descrition

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).

解析

与 3Sum 的思路一样。不同在于,我们现在希望找到距离 target 最近的数,参看代码。

code


#include <iostream>
#include <vector>
#include <algorithm>
#include <limits> using namespace std; class Solution{
public:
int threeSumClosest(vector<int>& nums, int target){
sort(nums.begin(), nums.end()); // ascending int min_gab = numeric_limits<int>::max();
int ans = target; for(int i=0; i<nums.size(); i++){
int target_local = target - nums[i];
int ileft = i + 1;
int iright = nums.size() - 1;
while(ileft < iright){ // two pointer searching
int sum = nums[ileft] + nums[iright];
if(sum == target_local) // right answer
return target;
if(sum < target_local) // move ileft to increase sum
ileft++;
else // sum > target_local
iright--; int gab = abs(sum - target_local);
if(gab < min_gab){
ans = sum + nums[i];
min_gab = gab;
}
}
} return ans; }
}; int main()
{
return 0;
}

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