Codility算法测验（三）

接上篇博文，第三个题目

A non-empty zero-indexed array A consisting of N positve integers is given. A pair of indices (P, Q), such that 0 <= P <= Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].
The equi-3 pair is a pair of indices (X, Y), such that 0 < X, X+1 < Y < N-1, and the sums of slices (0, X-1), (X+1, Y-1), (Y+1, N-1) are all equal. 
That, given a non-empty zero-indexed array A consisting of N integers, returns 1 if array A contains an equi-3 pair and 0 otherwise.
Assume that:
        N is an integer within the range [5..100,000];
        each element of array A is an integer within the range [1..10,000].
Complexity:
        expected worst-case time complexity is O(N);
        expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments). 

思考： 这个题目个人觉得最难的地方在时间复杂度是O(N), 因为要多次比较多个元素之和，嵌套循环似乎是避免不了的。后来看其他一些算法大神，学了个技巧，降低了复杂度。

public int solution(int[] A) {
        int sum = 0, leftsum = 0, midsum = 0, rightsum = 0;
        int n = A.length;
        int[] sums = new int[n];
        for (int i=0; i<n; i++) {
            sum += A;
            sums = sum;
        }
        int left = 1, right = n - 2;
        while (left+1 < right) {
            leftsum = sums[left-1];
            midsum = sums[right-1] - sums[left];
            rightsum = sums[n-1] - sums[right];
            if ((leftsum == midsum) && (midsum == rightsum)) {
                return 1;
            } else if (leftsum > rightsum) {
                right--;
            } else if (leftsum < rightsum){
                left++;
            } else {
                left++;
                right--;
            }
        }
        return 0;
    }