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).
AC code:
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int len = nums.size();
int front, tear, flag, sub=INT_MAX;
int ans, sum;
sort(nums.begin(), nums.end());
for (int i = 0; i < len; ++i) {
front = i+1;
tear = len-1;
while (front < tear) {
flag = sub;
sum = nums[i] + nums[front] + nums[tear];
if (sum > target) {
sub = min(sub, abs(sum - target));
if (flag != sub)
ans = target + sub;
tear--;
} else if (sum < target) {
sub = min(sub, abs(sum - target));
if (flag != sub)
ans = target - sub;
front++;
} else {
return target;
}
}
}
return ans;
}
};
Runtime: 12 ms, faster than 30.32% of C++ online submissions for 3Sum Closest.
原文：https://www.cnblogs.com/ruruozhenhao/p/9738639.html