题目
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).
分析
和15. 3Sum类似，甚至更加简单。先排序，再固定一个数，对剩下的数进行夹逼操作即可，时间复杂度O(n^2)。
实现
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int ans=nums[0]+nums[1]+nums[2];
sort(nums.begin(), nums.end());
for(int i=0; i<nums.size()-2; i++){
int begin=i+1, end=nums.size()-1;
while(begin<end){
int sum = nums[i]+nums[begin]+nums[end];
if(abs(sum-target)<abs(ans-target))
ans = sum;
if(nums[i]+nums[begin]+nums[end]<target)
begin++;
else
end--;
}
}
return ans;
}
};
思考
不知道为什么我的排名这么靠后，算法都差不多啊。我觉得一定是服务器的问题=_=
