[codility] EquiLeader解题报告

A non-empty zero-indexed array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:

    A[0] = 4    A[1] = 3    A[2] = 4    A[3] = 4    A[4] = 4    A[5] = 2

we can find two equi leaders:

• 0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
• 2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A consisting of N integers, returns the number of equi leaders.

For example, given:

    A[0] = 4    A[1] = 3    A[2] = 4    A[3] = 4    A[4] = 4    A[5] = 2

the function should return 2, as explained above.

Assume that:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

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).

Elements of input arrays can be modified.

解题思路：首先检查数组中是否存在leader。如果数组中不存在leader，必然不存在 equil leader。

证明：假设数组元素个数为n, 在下标S处将数组分割为两个子数组，如果S是equil leader, 那么数组中必然存在一个出现次数大于(S+1)/2 + (N-S-1)/2 >= (N)/2的数。

找到leader后，利用前缀计数组prefixCnt[0...N-1]保存子数组［0...i] (0 <= i <= N-1)中leader出现的次数。

那么对于index S(0 <= S <= N-1)，子数组 A[0...S]中leader个数为pfefixCnt[S]，子数组A[S+1...N-1] 中leader个数为数组A中leader出现总数减去prefixCnt[S]，只要保证这两个子数组中leader的出现次数都大于相应子数组长度的一半即可。

参考代码如下：

// 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    if(A.size() < 2)    return 0;        int leader = A[0];    int count = 1;    int size = A.size();        for(int i = 1; i < size; i++){        if(A[i] == leader){            count++;        }        else{            count --;            if(!count){                leader = A[i];                count = 1;            }        }    }        //valid whether the leader exists or not    count = 0;    vector<int> prefixCnt(size, 0);    for(int i = 0; i < size; i++){        if(A[i] == leader)  count++;        prefixCnt[i] = count;    }    if(count <＝ (size >> 1))   return 0;        int equi_leaders = 0;    for(int i = 0; i < size-1; i++){        if(prefixCnt[i] > ((i+1) >> 1) && count - prefixCnt[i] > ((size-i-1) >> 1) )            equi_leaders ++;    }        return equi_leaders;}