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描述：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.
思路1：Moore voting algorithm--每找出两个不同的element，就成对删除即count--，最终剩下的一定就是所求的。时间复杂度：O(n)
1 class Solution { 2 public: 3 int majorityElement(vector<int> &num) { 4 5 int elem = 0; 6 int count = 0; 7 8 for(int i = 0; i < num.size(); i++) { 9 10 if(count == 0) { 11 elem = num[i]; 12 count = 1; 13 } 14 else { 15 if(elem == num[i]) 16 count++; 17 else 18 count--; 19 } 20 21 } 22 return elem; 23 }
};
思路2：随机挑选一个元素，检查是否是多数元素。时间复杂度：Average：O(n)。期望查找次数 <2
1 class Solution { 2 public: 3 int majorityElement(vector<int> &num) { 4 5 int count = 0; 6 7 for(;;) { 8 if(num.size() == 1) 9 return num[0]; 10 else { 11 int i = rand() % (num.size() - 1); 12 for(int j = 0; j < num.size(); j++) { 13 if(num[j] == num[i]) 14 count++; 15 } 16 if(count > (num.size() / 2)) 17 return num[i]; 18 else { 19 count = 0; 20 continue; 21 } 22 } 23 } 24 }
};
附LeetCode建议解决方案：
LeetCode Problem: Majority Element查找多数元素
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原文地址：http://www.cnblogs.com/fanyabo/p/4178993.html
