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.
找到一堆数中出现次数>n/2 的那个数。
普通解法。可能用map 或者二分，但是仔细想一想可以用o（n）的解法。
其实就是求这堆数的众数。
class Solution { public: int majorityElement(vector<int>& nums) { int n = nums.size(); int can = nums[0]; int cnt = 1; for (int i = 1; i < n; ++i) { if (nums[i] == can) cnt++; else { cnt--; if (cnt == 0) can = nums[i],cnt = 1; } } return can; } };
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原文：http://www.cnblogs.com/pk28/p/7238990.html