Somewhat similar to the two problems:
First, a given containing 0, 1, 2, ..., n, the sequence number n, to find the 0 ... n does not appear in that number in the sequence.
Input: [3,0,1]
Output: 2
Input: [9,6,4,2,3,5,7,0,1]
Output: 8
Method a: Gauss formula: 1 + 2 + 3 + ... + n = (1 + n) * n / 2
public static int Solution(int[] arr) {
int len = arr.length;
int res = (len*(len+1))/2;
for (int i = 0; i < arr.length; i++) {
res -= arr[i];
}
return res;
}
Method two: Bitwise
public static int Solution(int[] arr) {
int res = 0;
for (int i = 0; i < arr.length; i++) {
res ^= arr[i]^i;
}
System.out.println("res="+res+",arr.length="+arr.length);
return res^arr.length;
}
The main function:
public static void main(String[] args) {
int[] arr = {8,6,9,2,3,5,7,0,1};
int res = Solution(arr);
System.out.println(res);
}
Second, all numbers in an array of length n's are in the range 0 to n-1, find a duplicate array of any number.
- Ideas:
- For solving this array element [0, n-1] the problem range, the element can be adjusted to a value i of i-th position.
In (2, 3, 1, 0, 2, 5), for example, to traverse the 4 position, this position is the number 2, but has a value of 2 in the second position, it is possible to know repeated 2 .
** time complexity is O (n), the spatial complexity is O (1).
public static int Solution(int[] arr) {
for (int i = 0; i < arr.length; i++) {
while(arr[i] != i) {
if(arr[i] == arr[arr[i]]) {
return arr[i];
}
swap(arr,arr[i],i);
}
}
return -1;
}
private static void swap(int[] arr, int m, int n) {
int temp = arr[m];
arr[m] = arr[n];
arr[n] = temp;
}
public static void main(String[] args) {
int[] arr = {2, 3, 1, 0, 2 ,5};
int res = Solution(arr);
System.out.println(res);
}
