【leetcode】Array—— Next Permutation（31）

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

代码：

public void nextPermutation(int[] nums) {
if(nums==null||nums.length<=1)
return;
int i=nums.length-2;
while (i>=0&&nums[i]>=nums[i+1])
i--;

if(i>=0){
int j=nums.length-1;
while(nums[j]<=nums[i])
j--;
swap(nums,i,j);
}

reverseSort(nums, i+1, nums.length-1);
}
public void swap(int[] num, int i, int j){
int temp=0;
temp=num[i];
num[i]=num[j];
num[j]=temp;
}

public void reverseSort(int[] num, int start, int end){
if(start>end)
return;
for(int i=start;i<=(end+start)/2;i++)
swap(num,i,start+end-i);
}

第一次尝试的时候，i 定位错误：

for(int i=nums.length-1;i>0;i--){
for(int j=i-1;j>=0;j--){
if(nums[i]>nums[j])
<span style="white-space:pre">		</span>//get the j
}
}

只需要判断相邻两位就能知道是符合从左往右是升序，找到“打破点”后，再找”交换点“