[LeetCode]Majority Element

标签：leetcode

题意: 找出一个数组中的主要元素,主要元素为出现次数大于[n/2]的元, 当然最笨的方法就是暴力,没式不知道时间行不行
思路1: 遍历数组,统计每个元素出现的次数,用HashMap统计每个元素出现的次数 时间复杂度O(N),空间复杂度 O(N)
代码1:

public class Solution {
       public int majorityElement(int[] num) {
        int majority = 0, currCount;
        int major = (int) Math.floor( num.length*1.0 / 2 );
        if(num.length == 1) return num[0];
        Map<Integer, Integer> m = new HashMap<Integer, Integer>();
        for(int i = 0; i< num.length; ++i){
            if(m.containsKey(num[i])){
                currCount = m.get(num[i]) + 1;
                if(currCount > major) return num[i];
                m.put(num[i], currCount);
            }else m.put(num[i],1);
        }

        return majority;
    }

}

    public int majorityElement(int[] num) {//排序
        int majority = 0, currCount;
        int major = (int) Math.floor( num.length*1.0 / 2 );
        if(num.length == 1) return num[0];

        Arrays.sort(num);

        return num[num.length/2];
    }

    public int majorityElement(int[] num) {//摩尔投票算法
        int majority = 0, currCounter = 0, candidate = 0;
        int minCounter = (int) Math.floor( num.length*1.0 / 2 );
        if(num.length == 1) return num[0];

        for(int i = 0; i < num.length; ++i){
            if(currCounter == 0){
                candidate = num[i];
                currCounter ++;
            }else {
                if(candidate == num[i]){
                    currCounter ++;
                }
                else currCounter --;
            }
            if(currCounter > minCounter)return candidate;
        }
        return candidate;
    }

[LeetCode]Majority Element

标签：leetcode

原文地址：http://blog.csdn.net/youmengjiuzhuiba/article/details/44887689