Majority Element [LeetCode 169]
2018-04-12 00:40
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Description
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.
Analysis
最简单的方法就是计算每个数出现的次数,当统计到一个数出现的次数大于n/2,那么就输出这个数。时间复杂度为O(n)。但由于我是选择的是Divide and Conquer这个topic,所以我尝试用分治的思想解决这个问题。
分治法的思想:
将大的问题分解为小的子问题,但子问题和大问题本质上为同一问题。
递归解决这些子问题。
将子问题的解答合并获得大问题的解答。
假设输入的是一个长度为n的vector,我们可以从中间把它分成两半,分别求出它们中出现次数最多的数。这两半也可以继续分解,分解到仅剩下一个数时停止。这就是一个把大问题分解为小的子问题的过程,我们要做的仍然是求出出现次数最多的数,但由于不断的分解,最后只剩下一个数时,这个数这个子问题中出现次数最多的数了。
Answer
class Solution { public: int majorityElement(vector<int>& nums) { return (majorpart(nums, 0, nums.size()-1)); } private: int majorpart(vector<int>& nums, int left, int right) { if (left == right) return nums[left]; int mid = left + ((right - left)/2); //递归解决子问题 int leftpart = majorpart(nums, left, mid); int rightpart = majorpart(nums, mid + 1, right); //cout << leftpart << ' ' << rightpart << endl; //将子问题的解答合并获得大问题的答案 if (leftpart == rightpart) return leftpart; else { int a = count(nums.begin() + left, nums.begin() + right + 1, leftpart); int b = count(nums.begin() + left, nums.begin() + right + 1, rightpart); return a>b? leftpart:rightpart; } } };
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