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.
题意：将原来的序列改成一个具有倒序的序列，如果已经是全部倒序，那么将它整个翻转过来就好。
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1
java代码：
public void nextPermutation(int[] nums) {
int index = nums.length - 1;
while (index > 0 && nums[index] <= nums[index - 1]) {
--index;
}
if (index == 0) {
Arrays.sort(nums);
return;
}
int second = Integer.MAX_VALUE, secondIndex = Integer.MAX_VALUE;
for (int i = nums.length - 1; i >= index - 1; --i) {
if (nums[i] > nums[index - 1] && nums[i] < second) {
second = nums[i];
secondIndex = i;
}
}
int tmp = nums[index - 1];
nums[index - 1] = nums[secondIndex];
nums[secondIndex] = tmp;
Arrays.sort(nums, index, nums.length);
}
