<p>Given an array of size <i>n</i>, find the majority element. The majority element is the element that appears <b>more than</b> <code>? n/2 ?</code> times.</p>

這道題本人也不會(huì)做，于是先查了一波代碼：

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

看到代碼后，不禁對(duì)這段代碼之作者產(chǎn)生了深深的敬仰之情~

其實(shí)這段代碼的本意是將兩個(gè)互不相同的數(shù)相互抵消，這樣子留下來的那個(gè)數(shù)就肯定是數(shù)量大于一半的那個(gè)數(shù)啦~

思路簡單易懂，復(fù)雜度也控制在<code>O(n)</code>,這段代碼真是美啊