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).题目大意是:在一个数组里寻找3个数字,这三个数字的和要和target最接近。
思路:先调用sort排序,然后取三个指针,p1从头开始遍历;p2指向p1的下一个,然后依次遍历;p3指向数组尾部;如果*p1+*p2+*p3>targrt,则应该--p3,相反则应该++p2,相等则最好,直接输出。
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
int ret;
bool first = false;
for(int i = 0; i < nums.size();i++){
int j = i+1;
int k = nums.size() -1;
while(j < k){
int sum = nums[i] + nums[j] + nums[k];
if(first){
ret = sum;
first = false;
}
else{
if(abs(sum - target) < abs(ret -target))
ret = sum;
}
if(ret == target)
return ret;
if(sum > target)
k--;
else
j++;
}
}
return ret;
}
};