[Codility] CountTriangles

A zero-indexed array A consisting of N integers is given. A triplet (P, Q, R) is triangular if it is possible to build a triangle with sides of lengths A[P], A[Q] and A[R]. In other words, triplet (P, Q, R) is triangular if 0 ≤ P < Q < R < N and:

• A[P] + A[Q] > A[R],
• A[Q] + A[R] > A[P],
• A[R] + A[P] > A[Q].

For example, consider array A such that:

  A[0] = 10    A[1] = 2    A[2] = 5
  A[3] = 1     A[4] = 8    A[5] = 12

There are four triangular triplets that can be constructed from elements of this array, namely (0, 2, 4), (0, 2, 5), (0, 4, 5), and (2, 4, 5).

Write a function:

int solution(vector<int> &A);

that, given a zero-indexed array A consisting of N integers, returns the number of triangular triplets in this array.

For example, given array A such that:

  A[0] = 10    A[1] = 2    A[2] = 5
  A[3] = 1     A[4] = 8    A[5] = 12

the function should return 4, as explained above.

Assume that:

• N is an integer within the range [0..1,000];
• each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

• expected worst-case time complexity is O(N2);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

给定正整数数组A,长度为N,下标从0开始，求(P,Q,R),满足0<=P<Q<R<N 并且 A[P] + A[Q] > A[R], A[Q] + A[R] > A[P], A[P] + A[R] > A[Q]的三元组个数。

数据范围 N [0..1000], 数组元素[1..10^9]。

要求复杂度 时间O(N ^ 2) ，空间 O(1)。

分析： 显然我们不能枚举……我们可以把数组排序 O(NlogN),甚至O(N^2)的排序都可以。然后还是枚举，只不过枚举两条较小的边A[x] , A[y], 然后我们考虑最大边A[z]，设想假设我们固定x, 当y变大时A[x] + A[y]也变大，我们需要A[x] + A[y] > A[z], y变大之前的那些z值现在依然也满足条件，所以我们只要接着上次满足条件的最大的z，继续循环就可以了。所以对于同一个x来说,y和z的变化都是O(N)的。总复杂度O(N^2)。

 // you can use includes, for example:
 #include <algorithm>

 // you can write to stdout for debugging purposes, e.g.
 // cout << "this is a debug message" << endl;

 int solution(vector<int> &A) {
     // write your code in C++11
     sort(A.begin(), A.end());
     int a, b, c;
     int res = ;
     for (a = ; a < (int)A.size() - ; ++a) {
         c = a + ;
         for (b = a + ; b < (int)A.size() - ; ++b) {
             for (c = max(c, b + ); c < A.size() && A[a] + A[b] > A[c]; ++c);
             res += c - b - ;
         }
     }
     return res;
 }