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题目链接：https://leetcode.com/problems/next-permutation/
题目：
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
思路：
见注释。。。。
算法：
public void nextPermutation(int[] nums) {
int pos = -1;
// 找到最后一个升序位置
for (int i = nums.length - 1; i >= 1; i--) {
if (nums[i] > nums[i - 1]) {
pos = i - 1;
break;
}
}
// 如果没有升序 如3 2 1 需要反转为1 2 3
if (pos < 0) {
nums = reverse(nums, 0, nums.length - 1);
return;
}
// 存在升序 找到pos之后最后一个大于pos的位置
// 举例 如1 2 5 4 3 1下一个排列应该是 1 3 1 2 4 5，
// 首先定位pos元素为2；找到元素3；两者交换：1 3 5 4 2 1；将5~1反转；
for (int i = nums.length - 1; i > pos; i--) {
if (nums[i] > nums[pos]) {
int tmp = nums[i];
nums[i] = nums[pos];
nums[pos] = tmp;
break;
}
}
// 将pos之后元素全部反转
nums = reverse(nums, pos + 1, nums.length - 1);
}
public int[] reverse(int nums[], int start, int end) {
while (start < end) {
int tmp = nums[start];
nums[start] = nums[end];
nums[end] = tmp;
start++;
end--;
}
return nums;
}
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原文地址：http://blog.csdn.net/yeqiuzs/article/details/51534178
