# Java O(n) time, O(1) space, swap number solution. Explanations

• Since the range of integers are from 1 to n, n is the size of array, and there is only one duplicate and one missing, you can realign the array in order using the following method.
For instance, [2,1,2,4] should become [1,2,2,4] after the swap. When you are swapping the second {2}, you find there is already a {2} in position, so it is the duplicate. And after all the rest numbers are in position, the index which the extra {2} takes is the missing number - 1.

``````public int[] findErrorNums(int[] nums) {
int[] ret = new int [2];
int i = 0;
while (i < nums.length) {
if (nums[i] == i+1) {
i++;
continue;
}
else {
if (nums[nums[i]-1] == nums[i]) {
ret[0] = nums[i];
i++;
continue;
}
else {
swap(nums[i]-1, i, nums);
}
}
}
for (int j = 0; j<nums.length; j++) {
if (nums[j] == ret[0] && j+1 != nums[j]) {
ret[1] = j+1;
}
}
return ret;
}
private void swap (int i, int j, int[] nums){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}``````

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