劍指offer---數組中重復的數字
阿新 • • 發佈:2017-08-04
blog n) true span inpu turn arr 元素 some
class Solution { public: // Parameters: // numbers: an array of integers // length: the length of array numbers // duplication: (Output) the duplicated number in the array number // Return value: true if the input is valid, and there are some duplications in the array number// otherwise false bool duplicate(int numbers[], int length, int* duplication) { if(length<=0||numbers==NULL) return false; //判斷每一個元素是否非法 for(int i=0;i<length;++i) { if(numbers[i]<=0||numbers[i]>length-1)return false; } for(int i=0;i<length;++i) { while(numbers[i]!=i) { if(numbers[i]==numbers[numbers[i]]) { *duplication = numbers[i]; return true; }//交換numbers[i]和numbers[numbers[i]] int temp = numbers[i]; numbers[i] = numbers[temp]; numbers[temp] = temp; } } return false; } };
劍指offer---數組中重復的數字