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题目：
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.
解法一：naive法
We can sort the array first, which takes time of nlog(n). Then scan once to find the longest consecutive substrings.
1 public class Solution { 2 public int majorityElement(int[] num) { 3 if(num.length==1){ 4 return num[0]; 5 } 6 7 Arrays.sort(num); 8 9 int prev=num[0]; 10 int count=1; 11 for(int i=1; i<num.length; i++){ 12 if(num[i] == prev){ 13 count++; 14 if(count > num.length/2) return num[i]; 15 }else{ 16 count=1; 17 prev = num[i]; 18 } 19 } 20 21 return 0; 22 } 23 }
解法二：Much Simpler
1 public int majorityElement(int[] num) { 2 if (num.length == 1) { 3 return num[0]; 4 } 5 6 Arrays.sort(num); 7 return num[num.length / 2]; 8 }
解法三：Linear Time Majority Vote Algorithm 很高级的样子
1 public int majorityElement(int[] nums) { 2 int result = 0, count = 0; 3 4 for(int i = 0; i<nums.length; i++ ) { 5 if(count == 0){ 6 result = nums[ i ]; 7 count = 1; 8 }else if(result == nums[i]){ 9 count++; 10 }else{ 11 count--; 12 } 13 } 14 15 return result; 16 }
reference：http://www.programcreek.com/2014/02/leetcode-majority-element-java/
http://www.cs.utexas.edu/~moore/best-ideas/mjrty/
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原文地址：http://www.cnblogs.com/hygeia/p/4645237.html
