Description
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1
Solution
One-way pass
真是历尽艰辛的一个solution...
class Solution {
public void nextPermutation(int[] nums) {
int i = nums.length - 2;
// find the longest subsquence in desc order
while (i >= 0 && nums[i + 1] <= nums[i]) {
i--;
}
if (i >= 0) {
int j = nums.length - 1;
// find the first pos larger than num[i], then swap
while (j >= 0 && nums[j] <= nums[i]) {
j--;
}
swap(nums, i, j);
}
// finally reverse [i+1, n-1]
reverse(nums, i + 1, nums.length - 1);
}
public void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
public void reverse(int[] nums, int begin, int end) {
while (begin < end) {
swap(nums, begin++, end--);
}
}
}
