Given an array
nums
of
n
integers and an integer
target
, find three integers in
nums
such that the sum is closest to
target
. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
非常开心的一道题目了,一次submission就accepted
和3sum类似用到了两个指针逐渐逼近
3 sum:https://blog.csdn.net/xueying_2017/article/details/79937063
water container:https://blog.csdn.net/xueying_2017/article/details/79934296
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(),nums.end()); int minDiff=INT_MAX; //最小的差值 int diff; int sign=1; //用来标记差值的符号 //int diffA;//差的绝对值 for(int i=0;i<nums.size();i++){ int front=i+1; int back=nums.size()-1; while(front<back){ diff=nums[i]+nums[front]+nums[back]-target; if(diff==0) return target; else if(diff>0){ if(diff<minDiff){ minDiff=diff; sign=1; } //如果差为正值的话,那么再小一点,差就有继续减小的可能 //和water container有异曲同工之妙 back--; } else if(diff<0){ if(abs(diff)<minDiff){ minDiff=abs(diff); sign=-1; } front++; } } } return target+sign*minDiff; } };
C++内的最大整数用常量 INT_MAX 表示
用abs( )求绝对值