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.

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

分析

分而治之。若nums只含一个数，则该数就是主元。若多于一个数，将nums分为两部分，若两部分都没有主元，则nums没有主元，若左右部分的主元相等，那么该数就是主元。若左右两边的主元不相等，扫描nums，统计该数的次数，若大于size/2，该数就是主元。

代码

class Solution {public:    int majorityElement(vector<int>& nums) {        int size=nums.size();        if (size==1)        return nums[0];        vector<int> left=vector<int>(nums.begin(),nums.begin()+size/2);        vector<int> right=vector<int>(nums.begin()+size/2,nums.end());        int left_major=majorityElement(left);        int right_major=majorityElement(right);        int tem;        if (left_major==-1&&right_major==-1) {        return -1;} else if (left_major==right_major) {return left_major;} else {if (left_major!=-1) {tem=left_major;int count = 0;for (int i = 0; i < size; ++i) {if (nums[i]==tem)count++;}if (count>size/2)return tem;}if (right_major!=-1) {tem=right_major;int count = 0;for (int i = 0; i < size; ++i) {if (nums[i]==tem)count++;}if (count>size/2)return tem;}return -1;}    }};