## 题目描述：

LeetCode 611. Valid Triangle Number

Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.

Example 1:

```Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
```

Note:

1. The length of the given array won't exceed 1000.
2. The integers in the given array are in the range of [0, 1000].

## Java代码：

``````public class Solution {

public int binarySearch(int[] nums, int start, int target) {
int left = start, right = nums.length - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] >= target) right = mid - 1;
else left = mid + 1;
}
return left;
}

public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int size = nums.length;
int ans = 0;
for (int i = 0; i < size - 2; i++) {
for (int j = i + 1; j < size - 1; j++) {
int k = binarySearch(nums, j + 1, nums[i] + nums[j]);
ans += k - j - 1;
}
}
return ans;
}

}
``````

## Java代码：

``````public class Solution {

public int triangleNumber(int[] nums) {
Arrays.sort(nums);
int size = nums.length;
int ans = 0;
for (int i = 0; i < size - 2; i++) {
if (nums[i] == 0) continue;
int k = i + 2;
for (int j = i + 1; j < size - 1; j++) {
while (k < size && nums[k] < nums[i] + nums[j]) k++;
ans += k - j - 1;
}
}
return ans;
}

}
``````

Pingbacks已关闭。

1. 书影网友 发布于 2017年6月15日 09:14 #

em... 还是比我写的好，佩服.
我的代码就要去处理一些小case
int j = i + 1, k = i + 2;
while (j &amp;lt; nums.length - 1 &amp;&amp; k &amp;lt; nums.length) {
k = Math.max(j + 1, k); //!!! 避免j,k重复
if (nums[i] + nums[j] &amp;lt;= nums[k]) {
j++;
} else {
res += k - j;
k++;
}
}

2. 书影网友 发布于 2017年8月6日 16:06 #

解法二是O（n^3）吧