https://leetcode.cn/problems/count-pairs-whose-sum-is-less-than-target/description/
给你一个下标从 0 开始长度为
n的整数数组nums和一个整数target,请你返回满足0 <= i < j < n且nums[i] + nums[j] < target的下标对(i, j)的数目。示例 1:
1
2
3
4
5
6
7 输入:nums = [-1,1,2,3,1], target = 2
>输出:3
解释:总共有 3 个下标对满足题目描述:
>- (0, 1) ,0 < 1 且 nums[0] + nums[1] = 0 < target
- (0, 2) ,0 < 2 且 nums[0] + nums[2] = 1 < target
- (0, 4) ,0 < 4 且 nums[0] + nums[4] = 0 < target
注意 (0, 3) 不计入答案因为 nums[0] + nums[3] 不是严格小于 target 。示例 2:
1
2
3
4
5
6
7
8
9
10
11
12
13 输入:nums = [-6,2,5,-2,-7,-1,3], target = -2
输出:10
解释:总共有 10 个下标对满足题目描述:
>- (0, 1) ,0 < 1 且 nums[0] + nums[1] = -4 < target
- (0, 3) ,0 < 3 且 nums[0] + nums[3] = -8 < target
>- (0, 4) ,0 < 4 且 nums[0] + nums[4] = -13 < target
- (0, 5) ,0 < 5 且 nums[0] + nums[5] = -7 < target
>- (0, 6) ,0 < 6 且 nums[0] + nums[6] = -3 < target
- (1, 4) ,1 < 4 且 nums[1] + nums[4] = -5 < target
>- (3, 4) ,3 < 4 且 nums[3] + nums[4] = -9 < target
- (3, 5) ,3 < 5 且 nums[3] + nums[5] = -3 < target
- (4, 5) ,4 < 5 且 nums[4] + nums[5] = -8 < target
- (4, 6) ,4 < 6 且 nums[4] + nums[6] = -4 < target提示:
1 <= nums.length == n <= 50-50 <= nums[i], target <= 50
先对数组进行排序,以保证当 nums[i] + nums[j] >= target 时任何后续的 j 都不用再遍历。
遍历
1 | class Solution { |
双指针
class Solution {
public int countPairs(List<Integer> nums, int target) {
Collections.sort(nums);
int ans = 0;
for (int i = 0, j = nums.size() - 1; i < j; i++) {
while (i < j && nums.get(i) + nums.get(j) >= target) j--;
ans += j - i;
}
return ans;
}
}