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[LeetCode] 259. 3Sum Smaller #259
Description
Given an array of n integers nums and a target , find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.
Example:
Input: _nums_ = [-2,0,1,3], and _target_ = 2
Output: 2
Explanation: Because there are two triplets which sums are less than 2:
[-2,0,1]
[-2,0,3]
Follow up: Could you solve it in O ( n 2) runtime?
这道题是 3Sum 问题的一个变形，让我们求三数之和小于一个目标值，那么最简单的方法就是穷举法，将所有的可能的三个数字的组合都遍历一遍，比较三数之和跟目标值之间的大小，小于的话则结果自增1，参见代码如下:
解法一：
// O(n^3)
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
int res = 0;
sort(nums.begin(), nums.end());
for (int i = 0; i < int(nums.size() - 2); ++i) {
int left = i + 1, right = nums.size() - 1, sum = target - nums[i];
for (int j = left; j <= right; ++j) {
for (int k = j + 1; k <= right; ++k) {
if (nums[j] + nums[k] < sum) ++res;
}
}
}
return res;
}
};
题目中的 Follow up 让我们在 O(n^2) 的时间复杂度内实现，那么借鉴之前那两道题 3Sum Closest 和 3Sum 中的方法，采用双指针来做，这里面有个 trick 就是当判断三个数之和小于目标值时，此时结果应该加上 right-left，因为数组排序了以后，如果加上 num[right] 小于目标值的话，那么加上一个更小的数必定也会小于目标值，然后将左指针右移一位，否则将右指针左移一位，参见代码如下：
解法二：
// O(n^2)
class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
if (nums.size() < 3) return 0;
int res = 0, n = nums.size();
sort(nums.begin(), nums.end());
for (int i = 0; i < n - 2; ++i) {
int left = i + 1, right = n - 1;
while (left < right) {
if (nums[i] + nums[left] + nums[right] < target) {
res += right - left;
++left;
} else {
--right;
}
}
}
return res;
}
};
Github 同步地址：
类似题目：
Two Sum Less Than K
参考资料：
https://leetcode.com/problems/3sum-smaller/
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