Description
Given an integer array nums of length n 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 1:
Input: nums = [-1,2,1,-4], target = 1 Output: 2 Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Example 2:
Input: nums = [0,0,0], target = 1 Output: 0 Explanation: The sum that is closest to the target is 0. (0 + 0 + 0 = 0).
Constraints:
3 <= nums.length <= 500-1000 <= nums[i] <= 1000-104 <= target <= 104
Code
3Sum 的變形版。
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int i = 0, l, r, n = nums.size();
int closestSum;
int distance = INT_MAX;
while(i < n) {
int l = i + 1, r = n - 1;
while(l < r) {
int Sum = nums[i] + nums[l] + nums[r];
if(Sum > target) {
if(abs(Sum - target) < distance) {
closestSum = Sum;
distance = abs(Sum - target);
}
r--;
} else if(Sum < target) {
if(abs(Sum - target) < distance) {
closestSum = Sum;
distance = abs(Sum - target);
}
l++;
} else if(Sum == target) {
closestSum = Sum;
return closestSum;
distance = 0;
int left = nums[l];
int right = nums[r];
while(l < r && nums[l] == left) l++;
while(l < r && nums[r] == right) r--;
}
}
while(i + 1 < n && nums[i] == nums[i + 1]) i++;
i++;
}
return closestSum;
}
};