LeetCode开心刷题第九天——17Letter Combinations of a Phone Number 18 4sum

最新推荐文章于 2024-06-28 09:31:18 发布

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最新推荐文章于 2024-06-28 09:31:18 发布

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原文链接：http://www.cnblogs.com/Marigolci/p/11073375.html

博客主要探讨两个问题，一是给定含特定范围数字的字符串，返回其可能代表的所有字母组合，介绍了BFS方法及时间、空间复杂度；二是给定整数数组和目标值，找出数组中所有和为目标值的唯一四数组合。

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17. Letter Combinations of a Phone Number

Medium

2241301FavoriteShare

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Special Point:
C11 is an effective tool.From now on I need to get familiar with C11.
Some usage in for loop need to be declare:

1.All These instance only want to declare one thing:
if you want to change value during C11 for loop,you need to use atuo&
Source of theory:
https://blog.youkuaiyun.com/hailong0715/article/details/54172848

    std::vector<int> vec {1,2,3,4,5,6,7,8,9,10};
    cout << "修改前" << endl;
    for (auto n :vec)
        std::cout << n++;
 
    cout << endl;
    cout << "修改后" << endl;
    for (auto j : vec)
        std::cout << j;

修改前
12345678910
修改后
12345678910

    std::vector<int> vec {1,2,3,4,5,6,7,8,9,10};
    cout << "修改前" << endl;
    for (auto& n :vec)
        std::cout << n++;
 
    cout << endl;
    cout << "修改后" << endl;
    for (auto j : vec)
        std::cout << j;

修改前
12345678910
修改后
234567891011

In this code swap means swap value between two vector

This way is BFS.Is easy than DFS

TIME COMPLEXITY:O(4^n)

SPACE COMPLEXITY:O(2*4^n) because it need two vector loop data through each other

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<vector>
#include<list>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;

class Solution{
public:

    vector<string> letterCombinations(string input)
    {
        if(input.empty()) return {};
        vector<vector<char>> temp(10);
        temp[0]={};
        temp[1]={};
        temp[2]={'a','b','c'};
        temp[3]={'d','e','f'};
        temp[4]={'g','h','i'};
        temp[5]={'j','k','l'};
        temp[6]={'m','n','o'};
        temp[7]={'p','q','r','s'};
        temp[8]={'t','u','v'};
        temp[9]={'w','x','y','z'};
        vector<string> res{""};
        for(char in:input)
        {
            vector<string> bridge;
            for(const string& s:res)
                for(char tp:temp[in-'0'])
                    bridge.push_back(s+tp);
            res.swap(bridge);
        }
        return res;
    }
};


int main()
{
    vector<string> res;
    string input;
    cin>>input;
    Solution s;
    res=s.letterCombinations(input);
    vector<string>::iterator iter=res.begin();
    for(iter;iter!=res.end();++iter)
    {
        cout<<*iter<<endl;
    }
    return 0;
}

18. 4Sum

Medium

1105219FavoriteShare

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

class Solution {
public:
    vector<vector<int>> fourSum(vector<int> &nums, int target) {
        vector<vector<int>> res;
        int n = nums.size();
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 3; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1; j < n - 2; ++j) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                int left = j + 1, right = n - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        vector<int> out{nums[i], nums[j], nums[left], nums[right]};
                        res.push_back(out);
                        while (left < right && nums[left] == nums[left + 1]) ++left;
                        while (left < right && nums[right] == nums[right - 1]) --right;
                        ++left; --right;
                    } else if (sum < target) ++left;
                    else --right;
                }
            }
        }
        return res;
    }
};