数组中重复的数字

题目描述

思路

``计数排序的思想，利用原数组的下标原地计数。``

代码

``````public class Solution {
// Parameters:
//    numbers:     an array of integers
//    length:      the length of array numbers
//    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
//                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
//    这里要特别注意~返回任意重复的一个，赋值duplication[0]
// Return value:       true if the input is valid, and there are some duplications in the array number
//                     otherwise false
public boolean duplicate(int numbers[],int length,int [] duplication) {
if(numbers == null || numbers.length == 0)    return false;
for(int i = 0; i < length; i++) {
int index = numbers[i];
if(index >= length) {
index -= length;
}
if(numbers[index] >= length) {
duplication[0] = index;
return true;
}
numbers[index] += length;
}
return false;
}
}``````

笔记

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