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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