Codility---Dominator

Task description

A zero-indexed array A consisting of N integers is given. The dominator of 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

class Solution { public int solution(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.

Solution
 
Programming language used: Java
Total time used: 1 minutes
 
Code: 01:47:41 UTC, java, final, score:  100
// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        int size = 0, leader = -1;
        for(int i=0; i<A.length;i++) {
            if(size == 0) {
                leader = A[i];
            }
            if(leader == A[i]) {
                size++;
            } else {
                size--;
            }
        }
        if(size == 0) return -1;
        int index = 0, count = 0;
        for(int i=0; i<A.length; i++) {
            if(leader == A[i]) {
                index = i;
                count++;
            }
        }
        if(count*2 <= A.length) return -1;
        return index;
    }
}


https://codility.com/demo/results/trainingK4VATM-6K8/
原文地址:https://www.cnblogs.com/samo/p/6984377.html