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3_ array face questions duplicate numbers
Thinking
- By one position determination, to the current position, or numbers [i] == i;! Or numbers [i] = i
- ! If numbers [i] = i, or numbers [numbers [i]] == numbers [i], returns to true;! Or numbers [numbers [i]] = numbers [i], index i and index exchange numbers [ i] corresponding to the value of
- After all the positions been traversed if not returns true, then there is no duplicate numbers, return false
the complexity
- Time complexity: Although the two cycles, but will be able to exchange a maximum of two numbers each in the correct position, the time complexity is O (n)
- Space complexity: no use of additional memory space, the space complexity is O (1)
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) {
// 健壮性判断
if(length <= 0)
return false;
for(int i = 0; i<length; i++){
if(numbers[i] > length - 1 || numbers[i] < 0)
return false;
}
// 正式判断, 逐个位置处理
for(int i = 0 ; i < length; i++){
while(numbers[i] != i){ // 直到把索引i的位置处理正确后才调出循环
int m = numbers[i];
if(m == numbers[m]){
duplication[0] = m;
return true;
}
// swap
int temp = numbers[i];
numbers[i] = numbers[m];
numbers[m] = temp;
}
}
return false;
}
}