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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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
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1 public class Solution { 2 public int majorityElement(int[] nums) { 3 if(nums == null) return 0; 4 Map<Integer,Integer> hm = new HashMap<Integer ,Integer>(); 5 for(int i= 0;i< nums.length;i++){ 6 if(hm.containsKey(nums[i])) hm.put(nums[i],hm.get(nums[i]) + 1); 7 else{ 8 hm.put(nums[i],1); 9 } 10 } 11 int halflen = nums.length/2; 12 int res = 0; 13 for(int key : hm.keySet()){ 14 if(hm.get(key) > halflen){ 15 res = key; 16 break; 17 } 18 } 19 return res; 20 } 21 }
方法2： 排序后，取中间的元素就是所求值。
方法3：转自：http://blog.csdn.net/feliciafay/article/details/45006661
从头到尾扫描一遍数组，记录当前的majority element的count。
1 // 259ms 2 public class Solution { 3 public int majorityElement(int[] num) { 4 if(num.length < 3) return num[0]; 5 int majority = num[0]; 6 int count = 1; 7 //1,1,1,1,2,1,3,1,2,2,2,2,2,2, 8 for (int i = 1; i < num.length; ++i) { 9 if (count == 0) { 10 majority = num[i]; 11 ++count; 12 } else if (num[i] == majority) { 13 ++count; 14 } else { 15 --count; 16 } 17 } 18 return majority; 19 } 20 }
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原文地址：http://www.cnblogs.com/guoguolan/p/5395968.html
