zoukankan      html  css  js  c++  java
  • [LintCode] 全排列

    递归实现:

    class Solution {
    public:
        /**
         * @param nums: A list of integers.
         * @return: A list of permutations.
         */
        vector<vector<int> > permute(vector<int> nums) {
            // write your code here
            vector<vector<int> > permutations;
            if (nums.empty()) return permutations;
            permutate(nums, 0, permutations);
            return permutations;
        }
    private:
        void permutate(vector<int> nums, int start, vector<vector<int> >& permutations) {
            if (start == nums.size()) {
                permutations.push_back(nums);
                return;
            }
            for (int i = start; i < (int)nums.size(); i++) {
                swap(nums[start], nums[i]);
                permutate(nums, start + 1, permutations);
            }
        }
    };

    非递归实现(基于nextPermutation):

     1 class Solution {
     2 public:
     3     /**
     4      * @param nums: A list of integers.
     5      * @return: A list of permutations.
     6      */
     7     vector<vector<int> > permute(vector<int> nums) {
     8         // write your code here
     9         vector<vector<int> > permutations;
    10         if (nums.empty()) return permutations;
    11         vector<int> copy(nums.begin(), nums.end());
    12         nextPermutation(nums);
    13         permutations.push_back(nums);
    14         while (nums != copy) {
    15             nextPermutation(nums);
    16             permutations.push_back(nums);
    17         }
    18         return permutations;
    19     }
    20 private:
    21     void nextPermutation(vector<int>& nums) {
    22         int k = -1, n = nums.size();
    23         for (int i = n - 2; i >= 0; i--) {
    24             if (nums[i] < nums[i + 1]) {
    25                 k = i;
    26                 break;
    27             }
    28         }
    29         if (k == -1) {
    30             reverse(nums.begin(), nums.end());
    31             return;
    32         }
    33         int l;
    34         for (int i = n - 1; i > k; i--) {
    35             if (nums[i] > nums[k]) {
    36                 l = i;
    37                 break;
    38             }
    39         }
    40         swap(nums[l], nums[k]);
    41         reverse(nums.begin() + k + 1, nums.end());
    42     }
    43 };
  • 相关阅读:
    记一次阿里云硬盘在线扩容
    大文件传输技巧-----split切割
    数据库迁移-------通过ibdata1文件和数据库文件迁移
    小技巧---------------vim 使用技巧 set paste 解决粘贴乱序问题
    webfrom 做项目的注意事项
    webform 复合控件
    wenfrom的简单控件和repeater控件
    分页功能 与 分类查询功能合并
    内置对象2
    简单的人员管理系统
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4625828.html
Copyright © 2011-2022 走看看