leetcode--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.

One : Sorting先排序，然后找major和count分别记录键值对count>n/2就返回major。

public class Solution {    public int majorityElement(int[] nums) {        Arrays.sort(nums);        int major = nums[0];        int count = 1;        for(int i = 1;i<nums.length;i++){            if(major==nums[i]){                count++;                if(count>nums.length/2){                    return major;                }            }else {                count = 1;                major = nums[i];            }        }        return major;    }}

Two：Sorting是one的优化

public int majorityElement1(int[] nums) {    Arrays.sort(nums);    return nums[nums.length/2];}

Three：Voting 投票机制。

public class Solution {    public int majorityElement(int[] nums) {               int major = nums[0];        int count = 1;        for(int i = 1;i<nums.length;i++){            if(count==0){                count++;                major = nums[i];            }else if(major == nums[i]) {                count++;            }else{                count--;            }        }        return major;    }}

Four:利用HashMap来操作Hashtable

public class Solution {    public int majorityElement(int[] nums) {               Map<Integer,Integer> myMap = new HashMap<Integer,Integer>();        int ret = nums[0];        for(int num : nums){            if(!myMap.containsKey(num)){                myMap.put(num,1);            }else{                myMap.put(num,myMap.get(num)+1);                if(myMap.get(num)>nums.length/2){                    ret = num;                    break;                }            }        }        return ret;    }}

Five:

// Bit manipulation public int majorityElement(int[] nums) {    int[] bit = new int[32];    for (int num: nums)        for (int i=0; i<32; i++)             if ((num>>(31-i) & 1) == 1)                bit[i]++;    int ret=0;    for (int i=0; i<32; i++) {        bit[i]=bit[i]>nums.length/2?1:0;        ret += bit[i]*(1<<(31-i));    }    return ret;}

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