31.Next Permutation

标签：swap   ext   perm   return   begin   class   end   public   next   

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int i, j, n = num.size();
        for (i = n - 2; i >= 0; --i) {
            if (num[i + 1] > num[i]) {
                for (j = n - 1; j > i; --j) {
                    if (num[j] > num[i]) break;
                }
                swap(num[i], num[j]);
                reverse(num.begin() + i + 1, num.end());
                return;
            }
        }
        reverse(num.begin(), num.end());
    }
};

class Solution {
public:
    void nextPermutation(vector<int>& nums) {int n = nums.size(), i = n - 2, j = n - 1;
        while (i >= 0 && nums[i] >= nums[i + 1]) --i;
        if (i >= 0) {
            while (nums[j] <= nums[i]) --j;
            swap(nums[i], nums[j]);
        }
        reverse(nums.begin() + i + 1, nums.end());
    }
};

31.Next Permutation

标签：swap   ext   perm   return   begin   class   end   public   next   

原文地址：https://www.cnblogs.com/smallredness/p/10673308.html