[Leetcode] 031. Next permutation (Medium) (C++/python)

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Github:https://github.com/illuz/leetcode

031. Next permutation (Medium) links：

Title: https://oj.leetcode.com/problems/next-permutation/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Find the next permutation of a sequence.

Analysis：

Can be lazy with the ' next_permutation ' in STL.

The specific algorithm is:

First, from the end of the beginning to look forward to two adjacent elements, so that the first element is I, the second element is II, and meet i<ii ;
Then, from the end of the beginning to search forward, find the first element greater than I, set it to J;
Then, I and J are swapped, and then all elements of the II and later are reversed.

Code：

C++:

Class Solution {public:    void Nextpermutation (vector<int> &num) {if (!num.size ()) Return;int idx = num.size ()-2;//1. Find out the last wrong orderwhile (idx >= 0 && num[idx] >= num[idx + 1]) idx--;//2. SWAPIF (idx >= 0) {int i = idx + 1;while (i < num.size () && num[i] > Num[idx]) i++;swap (num[i-1], Num[id x]);} 3. Reversereverse (Num.begin () + idx + 1, num.end ());}};

Python:

Class solution:    # @param num, a list of integer    # @return Nothing (void), does not return anything, modify Num IN-PL Ace instead.    def nextpermutation (self, num):        if not len (num):            return        idx = len (num)-2        # 1. "Find out" the last wrong or Der while        idx >= 0 and Num[idx] >= Num[idx + 1]:            idx-= 1        # 2. Swap        If idx >= 0:            i = idx +  1            While I < Len (num) and num[i] > Num[idx]:                i + = 1            num[i-1], num[idx] = Num[idx], num[i-1]        #  3. Reverse left        , right = idx + 1, len (num)-1 and left        <= right:            Num[left], num[right] = Num[right], Num[left] Left            + = 1 Right-            = 1

[Leetcode] 031. Next permutation (Medium) (C++/python)