[codility]Equi-leader

http://codility.com/demo/take-sample-test/equileader

一开始想到从左和右两边开始扫取众数，但求众数又要重新扫一遍，这样复杂度就是O(n^2)了。
此题的关键在于Equi-Leader必然是众数，否则不可能左边和右边都是众数。
所以先求出众数及其出现次数，再扫就行了。

// you can also use includes, for example:
// #include <algorithm>
int solution(vector<int> &A) {
    // write your code in C++98
    int x = A[0];
    int cnt = 0;
    for (int i = 1; i < A.size(); i++) {
        if (A[i] == x) {
            cnt++;
        }
        else if (cnt > 0) {
            cnt--;
        }
        else {
            cnt = 1;
            x = A[i];
        }
    }
    int total = 0;
    for (int i  = 0; i < A.size(); i++) {
        if (A[i] == x) total++;
    }
    if (total <= A.size() / 2) return 0;
    int ans = 0;
    int currentTotal = 0;
    for (int i = 0; i < A.size() - 1; i++) {
        if (A[i] == x)
            currentTotal++;
        if ((currentTotal > (i + 1) / 2) &&
            ((total - currentTotal) > (A.size() - i - 1) / 2)) {
            ans++;
        }
    }
    return ans;
}