[LeetCode#268]Missing Number

Problem:

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Analysis:

This kind of problem is very typical and quite tricky. 
Condition: the numbers in the array are distinct, and all of them fall into the range [0, n], and the array's size is n.
This condition means only one element would have no palce to reside in.

Tricky skill:
Scan from the first element, try to place all elements interwining with the number in the first element. 
for (int i = 0; i < nums.length; i++) {
    while (nums[i] != i && nums[i] < nums.length) {
        int temp = nums[i];
        nums[i] = nums[nums[i]];
        nums[temp] = temp;
    }
}
Unless the nums[i] = nums.length, we could always find the right position for nums[i].
Two stop cases:
a. nums[i] == i, which means nums[i] was placed at the right position.
b. nums[i] == nums.length, which means we could find no place to place the nums[i], otherwise exception appears. 

During the whole process, we place nums[i] to the right position, and exchange the nums[nums[i]] back to continue the placement process. 

After we finish the scan over the whole string, all placable elements were placed at their right positions. The unplacement element(if exist) was placed at the missing element's position. 

Two cases for answer:
1. the unplacement element was placed at the missing element's position.
for (int j = 0; j < nums.length; j++) {
    if (nums[j] != j)
        return j;
}
2. all elements are placable, and they were placed at their right position. 
return nums.length;



Errors:
while (nums[i] != i && i < nums.length) {
    int temp = nums[i];
    nums[i] = nums[nums[i]];
    nums[nums[i]] = temp;
}
The above code snippt could cause infinite loop.
Case [1, 0]
i = 0
temp = nums[0] (1);
nums[0] = nums[nums[0](1)] (0);
nums[nums[0]0] = temp;(1)
Do you see the loop?
Even the logic in swapping is right, but the index<nums[i]> we use was changed in the middle!!!
nums[i] was changed at: nums[i] = nums[nums[i]];
This is very special, a way to fix it is to record nums[i]'s value
while (nums[i] != i && nums[i] < nums.length) {
    int temp = nums[i];
    nums[i] = nums[nums[i]];
    nums[temp] = temp;
}

Solution:

public class Solution {
    public int missingNumber(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        for (int i = 0; i < nums.length; i++) {
            while (nums[i] != i && nums[i] < nums.length) {
                int temp = nums[i];
                nums[i] = nums[nums[i]];
                nums[temp] = temp;
            }
        }
        for (int j = 0; j < nums.length; j++) {
            if (nums[j] != j)
                return j;
        }
        return nums.length;
    }
}