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(n2) 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 = ;
sort(nums.begin(), nums.end());
for (int i = ; i < int(nums.size() - ); ++i) {
int left = i + , right = nums.size() - , sum = target - nums[i];
for (int j = left; j <= right; ++j) {
for (int k = j + ; 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() < ) return ;
int res = , n = nums.size();
sort(nums.begin(), nums.end());
for (int i = ; i < n - ; ++i) {
int left = i + , right = n - ;
while (left < right) {
if (nums[i] + nums[left] + nums[right] < target) {
res += right - left;
++left;
} else {
--right;
}
}
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/259
类似题目:
Two Sum Less Than K
参考资料: