第二周：[LeetCode]169. Majority Element

[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.

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方法一：分治法: 将数组拆成两半, 分别找出前一半的众数和后一半的众数，若这两个众数相等，则全局众数就为改数，若不相等，则出现次数较多的众数为全局众数。时间复杂度为T(n) = T(n/2) + 2n = O(n log n).

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class Solution {public:    int majorityElement(vector<int>& nums) {        return divide_compare(nums,0,nums.size()-1);    }    int divide_compare(vector<int>& nums,int left,int right){        if (left == right)            return nums[left];        int mid = left + (right - left)/2;        int left_m = divide_compare(nums, left, mid);        int right_m = divide_compare(nums, mid+1, right);        if (left_m == right_m)            return left_m;        int left_m_count = count(nums.begin()+left,nums.begin()+right+1,left_m);        int right_m_count = count(nums.begin()+left, nums.begin()+right+1, right_m);        return left_m_count > right_m_count ? left_m: right_m;    }}; 

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方法二：Moore投票算法（相当于简单实现的分治法）: 维护一个当前的候选众数和一个初始为0的计数器。遍历数组时，当前值为x，如果计数器是0, 将候选众数置为x,并将计数器加一，如果计数器非0, 若x与候选众数相等，计数器加1，否则计数器减1（相当于不同的数互相抵消）。遍历完数组后, 当前的候选众数就是所求众数。时间复杂度 = O(n)。

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    int majorityElement(vector<int>& nums) {         int i,candidate=nums[0],count=1;        for(i=1; i<nums.size();i++){            if(count == 0){                candidate = nums[i];                count++;                continue;            }            if(candidate == nums[i])               count ++;            else               count --;        }        return candidate;    }};