[LeetCode]: 169: Majority Element

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.

思路：直接查找，使用arraylist记录结果，出现超过一半的数就返回

代码：

public class Solution {public static int bExist(ArrayList<int[]> nums,int intInput) {for(int i = 0;i<nums.size();i++){if( nums.get(i)[0]== intInput){return i;}}return -1;}public static int majorityElement(int[] nums) {ArrayList<int[]> arrResult  = new  ArrayList<int[]>();for(int i = 0;i<nums.length;i++){int intPos = bExist(arrResult,nums[i]);if( intPos == -1){int[] arrTemp = {nums[i],0};arrResult.add(arrTemp);}else{arrResult.get(intPos)[1] += 1; if(arrResult.get(intPos)[1] > (nums.length/2)){return arrResult.get(intPos)[0];}}}int intTemp = 0;int intPos = 0;for(int i = 0;i<arrResult.size();i++ ){if(intTemp <= arrResult.get(i)[1]){intTemp =arrResult.get(i)[1];intPos = i;}}return arrResult.get(intPos)[0];}
}

网上高人的思路：每找出两个不同的element，则成对删除。最终剩下的一定就是所求的。

代码：

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

转载于:https://www.cnblogs.com/savageclc26/p/4830970.html

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