Codility-Dominator

最新推荐文章于 2024-09-16 17:53:19 发布

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最新推荐文章于 2024-09-16 17:53:19 发布

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本文链接：https://blog.youkuaiyun.com/liushuying668/article/details/38044113

本文介绍了一种高效查找数组中出现次数超过一半的主导元素的方法，并提供了详细的算法实现步骤及示例代码。

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Task description

A zero-indexed array A consisting of N integers is given. The dominatorof array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

A[0] = 3    A[1] = 4    A[2] =  3
A[3] = 2    A[4] = 3    A[5] = -1
A[6] = 3    A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

int solution(const vector<int> &A);

that, given a zero-indexed array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

Assume that:

N is an integer within the range [0..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, given array A such that

A[0] = 3    A[1] = 4    A[2] =  3
A[3] = 2    A[4] = 3    A[5] = -1
A[6] = 3    A[7] = 3

the function may return 0, 2, 4, 6 or 7, as explained above.

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

此题放在lesson 6--leader，在其阅读材料中讲了选取leader的方法，我们只要在选取leader的同时记录下该元素下标即可。

选取leader的思路如下：

注意到如果序列a1~~an中含有leader，那么在移除两个不相等的元素后，剩余序列与原序列有相同的leader。

建立一个空栈，将a1~~an依次压入，每次压入新元素后检查栈顶的两个元素是否相等，如不等，则将这两个元素移除。最后，如果该序列中确实含有leader，则栈中剩下的元素就是leader。

事实上该序列中可能不含leader，故最后要检验一下栈中剩余元素是否为leader。

代码如下：

// you can use includes, for example:
// #include 

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

int solution(const vector &A) {
    // write your code in C++11
    int value = 0;
    int size = 0;
    int pos = -1;
    for(unsigned int i=0;iA.size()/2)
    return pos;
    else
    return -1;
}