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劍指offer---數組中重復的數字

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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