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【LeetCode】Majority Element

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.

 

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。

可扩展到⌊ n/k ⌋的情况,每k个不同的element进行成对删除。

class Solution {public:    int majorityElement(vector<int> &num) {        int nTimes = 0;        int candidate = 0;        for(int i = 0; i < num.size(); i ++)        {            if(nTimes == 0)            {                candidate = num[i];                nTimes = 1;            }            else            {                if(candidate == num[i])                    nTimes ++;                else                    nTimes --;            }        }        return candidate;    }};

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【LeetCode】Majority Element