LeetCode 229: Majority Element II

Array LeetCode

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.
Hint:

1. How many majority elements could it possibly have?
2. Do you have a better hint? Suggest it!

题意

思路

这道题让我们求出现次数大于n/3的众数，而且限定了时间和空间复杂度，那么就不能排序，也不能使用哈希表，这么苛刻的限制条件只有一种方法能解了，那就是摩尔投票法 Moore Voting，这种方法在之前那道题Majority Element 求众数中也使用了。题目中给了一条很重要的提示，让我们先考虑可能会有多少个众数。那么有了这个信息，我们使用投票法的核心是找出两个候选众数进行投票，需要两遍遍历，第一遍历找出两个候选众数，第二遍遍历重新投票验证这两个候选众数是否为众数即可，选候选众数方法和前面那篇Majority Element 求众数一样，由于之前那题题目中限定了一定会有众数存在，故而省略了验证候选众数的步骤，这道题却没有这种限定，即满足要求的众数可能不存在，所以要有验证。代码如下：

代码

C语言版本：

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* Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* (int* nums, int numsSize, int* returnSize) { int* res = (int*)malloc(sizeof(int)*numsSize); int firstNum = 0, secondNum = 0, countFN = 0, countSN = 0; for(int i=0; i<numsSize; i++){ if (nums[i] == firstNum){ countFN++; }else if (nums[i] == secondNum){ countSN++; }else if (countFN == 0){ firstNum = nums[i]; countFN = 1; }else if (countSN == 0){ 大专栏  LeetCode 229: Majority Element II secondNum = nums[i]; countSN = 1; }else{ countFN--; countSN--; } } countFN = countSN = 0; for (int i=0; i<numsSize; i++) { if (nums[i] == firstNum){ ++countFN; } else if (nums[i] == secondNum){ ++countSN; } } if (countFN > numsSize/3){ res[0] = firstNum; } if (countSN > numsSize/3){ res[1] = secondNum; } *returnSize = numsSize; return res; }

C++版本：

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class Solution { public: vector<int> majorityElement(vector<int>& nums) { vector<int> res; int m = 0, n = 0, cm = 0, cn = 0; for (auto &a : nums) { if (a == m) ++cm; else if (a ==n) ++cn; else if (cm == 0) m = a, cm = 1; else if (cn == 0) n = a, cn = 1; else --cm, --cn; } cm = cn = 0; for (auto &a : nums) { if (a == m) ++cm; else if (a == n) ++cn; } if (cm > nums.size() / 3) res.push_back(m); if (cn > nums.size() / 3) res.push_back(n); return res; } };