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1 class Solution { 2 public: 3 // Parameters: 4 // numbers: an array of integers 5 // length: the length of array numbers 6 // duplication: (Output) the duplicated number in the array number 7 // Return value: true if the input is valid, and there are some duplications in the array number 8 // otherwise false 9 bool duplicate(int numbers[], int length, int* duplication) { 10 if(length < 2) 11 return 0; 12 map<int,int> mm; 13 for (int i = 0; i< length ; ++i) 14 { 15 if (mm.count(numbers[i])==0) 16 { 17 mm[numbers[i]] = 1; 18 } 19 else 20 { 21 ++mm[numbers[i]]; 22 } 23 } 24 25 for(map<int,int>::iterator it = mm.begin() ;it != mm.end() ; ++it) 26 { 27 if (it->second > 1) 28 { 29 * duplication = it->first; 30 return 1; 31 } 32 } 33 return 0; 34 } 35 };
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原文地址:http://www.cnblogs.com/xiaoyesoso/p/5161840.html