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1. Use memory :
In the question, it already stated that there must be a majority element. So discard empty situation.
Also it needs the number, not the index.
1 class Solution { 2 public: 3 int majorityElement(vector<int> &num) { 4 int len = num.size(), result = 0 ,rec = 0; 5 unordered_map<int, int> list; 6 for (int i = 0; i < len; i++) { 7 if (list[num[i]]) list[num[i]]++; 8 else list[num[i]] = 1; 9 if (rec < list[num[i]]) { 10 rec = list[num[i]]; 11 result = num[i]; 12 } 13 } 14 return result; 15 } 16 };
2. With out memory:
Notes:
1. initialize the record after sorting. Because the ordering has been changed.
2. As question mentioned, majority over (n/2) times. Just use len/2 as the condition.
3. Return value is num[i-1] not num[i]. Because the num[i] is not the current value. Also we can put return current;
1 class Solution { 2 public: 3 int majorityElement(vector<int> &num) { 4 sort(num.begin(), num.end()); 5 int len = num.size(), rec = 1, current = num[0]; 6 for (int i = 1; i < len; i++) { 7 if (current != num[i]) { 8 if (rec > len/2) { 9 return num[i-1]; 10 } 11 current = num[i]; 12 rec = 1; 13 } else { 14 rec++; 15 } 16 } 17 return num[len-1]; 18 } 19 };
LeetCode – Refresh – Majority Element
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原文地址：http://www.cnblogs.com/shuashuashua/p/4352732.html
