LeetCode 169. Majority Element

问题:

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.

 

分析:

 

1 计数?

2 HashMap

 

first try:

import java.util.Map;
import java.util.HashMap;

class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> elementCountMap = new HashMap<Integer, Integer>();
        
        for(int i=0; i<nums.length; i++) {
            if(elementCountMap.containsKey(nums[i])) {
                int count = elementCountMap.get(nums[i]);
                elementCountMap.put(nums[i], count++);
            } else {
                elementCountMap.put(nums[i], 1);
            }
        }//
        
        int majorElement = 0;
        for(Map.Entry entry : elementCountMap.entrySet()) {
            if ((nums.length/2)<(Integer)entry.getValue()) {
                majorElement = (Integer)entry.getKey();
            }
        }
        
        return majorElement;
    }
}

 

 

result:

Submission Result: Wrong Answer 
Input: [2,2]
Output: 0
Expected: 2

 

anylysis:

找了很久也找不到错误。甚至还处处println debug,后来发现了犯了低级错误! count++啊啊啊啊

 

second try:

import java.util.Map;
import java.util.HashMap;

class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> elementCountMap = new HashMap<Integer, Integer>();
        
        for(int i=0; i<nums.length; i++) {
            if(elementCountMap.containsKey(nums[i])) {
                int count = elementCountMap.get(nums[i]);
                elementCountMap.put(nums[i], ++count);
            } else {
                elementCountMap.put(nums[i], 1);
            }
        }//
        
        int majorElement = 0;
        for(Map.Entry entry : elementCountMap.entrySet()) {
            if ((nums.length/2)<(Integer)entry.getValue()) {
                majorElement = (Integer)entry.getKey();
            }
        }
        
        return majorElement;
    }
}

 

result:

 

 

third try:

//import java.util.Map;
import java.util.HashMap;

class Solution {
    public int majorityElement(int[] nums) {
        HashMap<Integer, Integer> elementCountMap = new HashMap<Integer, Integer>();
        
        for(int i=0; i<nums.length; i++) {
            if(elementCountMap.containsKey(nums[i])) {
                int count = elementCountMap.get(nums[i]);
                elementCountMap.put(nums[i], ++count);
            } else {
                elementCountMap.put(nums[i], 1);
            }
        }//
        
        int majorElement = 0;
        for(Map.Entry entry : elementCountMap.entrySet()) {
            if ((nums.length/2)<(Integer)entry.getValue()) {
                majorElement = (Integer)entry.getKey();
            }
        }
        
        return majorElement;
    }
}

 

result:

分析:

怎么效率还变差了。。。

 

try again:

//import java.util.Map;
import java.util.HashMap;

class Solution {
    public int majorityElement(int[] nums) {
        HashMap<Integer, Integer> elementCountMap = new HashMap<Integer, Integer>();
        
        for(int i=0; i<nums.length; i++) {
            if(elementCountMap.containsKey(nums[i])) {
                int count = elementCountMap.get(nums[i]);
                elementCountMap.put(nums[i], ++count);
            } else {
                elementCountMap.put(nums[i], 1);
            }
        }//
        
        int majorElement = 0;
        for(Map.Entry entry : elementCountMap.entrySet()) {
            if ((nums.length/2)<(Integer)entry.getValue()) {
                majorElement = (Integer)entry.getKey();
                return majorElement;
            }
        }
        
        return majorElement;
    }
}

 


result:

精妙算法BM投票算法:

分析:

是流算法的一个prototypical example:

In computer science, streaming algorithms are algorithms for processing data streams in which the input is presented as a sequence of items and can be examined in only a few passes (typically just one). In most models, these algorithms have access to limited memory (generally logarithmic in the size of and/or the maximum value in the stream). They may also have limited processing time per item.

try:

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

 

result:

 

conclusion:

 

posted @ 2018-04-21 15:12  Zhao_Gang  阅读(96)  评论(0编辑  收藏  举报