# LeetCode 169. Majority Element

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.

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:

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:

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:

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