LeetCode 31 Next Permutation (C,C++,Java,Python)

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Problem:

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:採用倒序尋找的方法，尋找到第一個遞減的數字，比如2321，第一個遞減的數字是2，然後在將這個遞減數字後邊的比它大的最小的數字互換，2321就為3221，然後將後邊的數字升序排列，這樣就是了

題目大意：給定一個數組，重新排列這些數字，使得排列好的數字比原來的數字大，並且這個數字是比原來數字大的集合中最小的
Java原始碼（365ms）：

public class Solution {    public void nextPermutation(int[] nums) {        int len=nums.length,i=len-1,j,tmp;        while(i>0 && nums[i]<=nums[i-1])i--;        if(i>0){            j=len-1;            while(nums[j]<=nums[i-1])j--;            tmp=nums[j];            nums[j]=nums[i-1];            nums[i-1]=tmp;        }        j=len-1;        while(i<j){            tmp=nums[i];            nums[i]=nums[j];            nums[j]=tmp;            i++;j--;        }    }}

C語言原始碼（14ms）：

void nextPermutation(int* nums, int numsSize) {    int j,i=numsSize-1,tmp;    while(i>0 && nums[i]<=nums[i-1])i--;    if(i!=0){        j=numsSize-1;        while(j>=i && nums[j]<=nums[i-1])j--;        tmp=nums[j];        nums[j]=nums[i-1];        nums[i-1]=tmp;    }    j=numsSize-1;    while(i<j){        tmp=nums[i];        nums[i]=nums[j];        nums[j]=tmp;        i++;j--;    }}

C++原始碼（12ms）：

class Solution {public:    void nextPermutation(vector<int>& nums) {        int j,len=nums.size(),i=len-1,tmp;        while(i>0 &&  nums[i]<=nums[i-1])i--;        if(i>0){            j=len-1;            while(j>=i && nums[j]<=nums[i-1])j--;            tmp=nums[i-1];            nums[i-1]=nums[j];            nums[j]=tmp;        }        j=len-1;        while(i<j){            tmp=nums[i];            nums[i]=nums[j];            nums[j]=tmp;            i++;j--;        }    }};

Python原始碼（136ms）：

class Solution:    # @param {integer[]} nums    # @return {void} Do not return anything, modify nums in-place instead.    def nextPermutation(self, nums):        length=len(nums);i=length-1        while i>0 and nums[i]<=nums[i-1]:i-=1        if i>0:            j=length-1            while nums[j]<=nums[i-1]:j-=1            tmp=nums[j]            nums[j]=nums[i-1]            nums[i-1]=tmp        j=length-1        while i<j:            tmp=nums[i]            nums[i]=nums[j]            nums[j]=tmp            i+=1;j-=1

LeetCode 31 Next Permutation (C,C++,Java,Python)