# Triangle Count

## Question

Given an array of integers, how many three numbers can be found in the array, so that we can build an triangle whose three edges length is the three numbers that we find?

Example

Given array S = [3,4,6,7], return 3. They are:

``````[3,4,6]
[3,6,7]
[4,6,7]
``````

Given array S = [4,4,4,4], return 4. They are:

``````[4(1),4(2),4(3)]
[4(1),4(2),4(4)]
[4(1),4(3),4(4)]
[4(2),4(3),4(4)]
``````

## Analysis

1. 对数组排序，按照O(nlogn)计
2. 对数组下标循环，则内部转化为一个two sum II问题，即寻找 S[j] + S[k] > S[i]有多少组，因为数组已排序，则可以使用two pointers的方法
3. 对于每一个i，初始化left = 0, right = i - 1，如果有一个满足S[left] + S[right] > S[i], 那么对于left ~ right - 1 同样也满足，因此计入right-left到最终count中

## Solution

``````public class Solution {
/**
* @param S: A list of integers
* @return: An integer
*/
public int triangleCount(int S[]) {
Arrays.sort(S);
int count = 0;
for (int i = 0; i < S.length; i++) {
int left = 0;
int right = i - 1;
while (left < right) {
if (S[left] + S[right] > S[i]) {
// The edge from S[left] to S[right - 1] will also form a triangle
count += right - left;
right--;
} else {
left++;
}
}
}
return count;
}
}
``````