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.
題意：將原來的序列改成一個(gè)具有倒序的序列，如果已經(jīng)是全部倒序肠阱，那么將它整個(gè)翻轉(zhuǎn)過來就好呆馁。
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1
java代碼：
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);
}
