lintcode-medium-Next Permutation

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Given a list of integers, which denote a permutation.

Find the next permutation in ascending order.

Notice

The list may contains duplicate integers.

Example

For

[1,3,2,3]

, the next permutation is

[1,3,3,2]

For

[4,3,2,1]

, the next permutation is

[1,2,3,4]

public class Solution {
/**
* @param nums: an array of integers
* @return: return nothing (void), do not return anything, modify nums in-place instead
*/
public int[] nextPermutation(int[] nums) {
// write your code here

if(nums == null || nums.length <= 1)
return nums;

int i = nums.length - 1;

while(i > 0 && nums[i - 1] >= nums[i])
i--;

if(i == 0){
reverse(nums, 0, nums.length - 1);
return nums;
}

i--;

int j = i + 1;

while(j < nums.length && nums[j] > nums[i])
j++;

j--;

swap(nums, i, j);
reverse(nums, i + 1, nums.length - 1);

return nums;
}

public void swap(int[] nums, int i, int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;

return;
}

public void reverse(int[] nums, int start, int end){
while(start < end){
swap(nums, start, end);
start++;
end--;
}

return;
}

}