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  • Codility---EquiLeader

    Task description

    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:

    class Solution { public int solution(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.

    Solution
     
    Programming language used: Java
    Total time used: 26 minutes
     
    Code: 02:26:16 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, count=0, fcnt=0, res=0;
            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 0;
            for(int i=0; i<A.length; i++) {
                if(leader == A[i]) {
                    count++;
                }
            }
            for(int i=0; i<A.length; i++) {
                if(leader == A[i] ) {
                    fcnt++;
                }
                if((i+1) < fcnt*2 && (A.length-i-1) < (count-fcnt)*2) {
                    res++;
                }
            }
            return res;
        }
    }


    https://codility.com/demo/results/trainingKHWPS7-27V/
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  • 原文地址:https://www.cnblogs.com/samo/p/6984454.html
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