Majority Element

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Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

#include<stdio.h>void sort(int num[], int low, int high){    if(low < high){        int val = num[low], i = low, j = high;        while(i < j){            while(num[j] >= val && i < j) j--;            if(i < j) num[i++] = num[j];             while(num[i] < val && i < j) i++;            if(i < j) num[j--] = num[i];        }        num[i] = val;        sort(num, low, i-1);        sort(num, i+1, high);    }}//先排序再找,时间复杂度太高,超时了int majorityElement1(int num[], int n) {    sort(num, 0, n-1);    return num[n/2];}
//这个方法accepted了
int majorityElement(int num[], int n) {    int result = 0;    int time = 0 ,i;    for(i = 0; i < n; i++) {        if(time == 0) {            result = num[i];            time++;        }        else {            if(num[i] == result) time++;            else time--;        }    }    return result;}void main(){    int num[] = {3,3,3,3,3,3,3,1,1,1,22,122,11,15,1,2,3,3,3,3,4};    int n = 21, i;    printf("%d\n", majorityElement(num,n));}




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