Leetcode C++ 算法

Leetcode题解<四>

算法分析与设计课后作业

Posted by tomylee on December 3, 2017

No.1

题目链接：Combination Sum

  • 题目描述：

    Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

    The same repeated number may be chosen from C unlimited number of times.

    Note:

    • All numbers (including target) will be positive integers.
    • The solution set must not contain duplicate combinations.

    For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is:

    [
[7],
[2, 2, 3]
]

(1)思路：使用深度优先的方法搜索，并且用一个vector记录向量，找到合适的向量时将它保存在结果中并进行回溯操作。

(2)代码：

class Solution {
public:
   vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > result;
        vector<int> path;
        sort(candidates.begin(),candidates.end());
        helper(candidates,0,0,target,path,result);
        return result;
    }

    void helper(vector<int> &nums,int pos,int base,int target,vector<int>& path,vector<vector<int> > & result)
    {
        if(base==target)
        {
            result.push_back(path);
            return ;
        }
        if(base>target)
            return ;
        for(int i=pos;i<nums.size();i++)
        {
            path.push_back(nums[i]);
            helper(nums,i,base+nums[i],target,path,result);
            path.pop_back();
        }
    }
};

(3)提交结果：

这里写图片描述

(4)测试代码：

#include<iostream>

#include<string>

#include<algorithm>

#include<vector>

using namespace std;

class Solution {
public:
   vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        vector<vector<int> > result;
        vector<int> path;
        sort(candidates.begin(),candidates.end());
        helper(candidates,0,0,target,path,result);
        return result;
    }

    void helper(vector<int> &nums,int pos,int base,int target,vector<int>& path,vector<vector<int> > & result)
    {
        if(base==target)
        {
            result.push_back(path);
            return ;
        }
        if(base>target)
            return ;
        for(int i=pos;i<nums.size();i++)
        {
            path.push_back(nums[i]);
            helper(nums,i,base+nums[i],target,path,result);
            path.pop_back();
        }
    }
};

int main(){
    int n; 
    cout<<"请输入向量长度："; 
    cin>>n;
    vector<int> nums;
    int a;
    for(int i=0;i<n;i++){
        cin>>a;
        nums.push_back(a);
    }
    int target;
    cout<<"请输入target:";
    cin>>target;
    Solution solution;
    vector<vector<int> > result=solution.combinationSum(nums, target);
    int size = result.size();
    for(int i=0;i<size;i++){
        int size1=result[i].size();
        cout<<"[";
        int j;
        for(j=0;j<size1-1;j++){
            cout<<result[i][j]<<", ";   
        }
        cout<<result[i][j]<<"]"<<endl;
    }
}

No.2

题目链接：Combination Sum II

  • 题目描述：

    Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

    Each number in C may only be used once in the combination.

    Note:

    • All numbers (including target) will be positive integers.
    • The solution set must not contain duplicate combinations.

    For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8, A solution set is:

    [
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

(1)思路：递归回溯，但注意在同一层递归树中，如果某元素已经处理并进入下一层递归，那么与该元素相同的值就应该跳过,否则将出现重复;例如：1,1,2,3 如果第一个1已经处理并进入下一层递归1,2,3。那么第二个1就应该跳过，因为后续所有情况都已经被覆盖掉。 另外，相同元素第一个进入下一层递归，而不是任意一个。例如：1,1,2,3。如果第一个1已经处理并进入下一层递归1,2,3，那么两个1是可以同时成为可行解的，而如果选择的是第二个1并进入下一层递归2,3，那么不会出现两个1的解了。

(2)代码：

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > ret;
        vector<int> cur;
        Helper(ret, cur, candidates, target, 0);
        return ret;
    }
    void Helper(vector<vector<int> > &ret, vector<int> cur, vector<int> &candidates, int target, int position){
        if(target == 0)
            ret.push_back(cur);
        else{
            for(int i = position; i < candidates.size() &&candidates[i] <= target; i ++){
                if(i != position && candidates[i] == candidates[i-1])
                    continue;
                cur.push_back(candidates[i]);
                Helper(ret, cur, candidates, target-candidates[i], i+1);
                cur.pop_back();
            }
        }
    }

};

(3)提交结果：

这里写图片描述

(4)测试代码：

#include<iostream>

#include<string>

#include<algorithm>

#include<vector>

using namespace std;

class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > ret;
        vector<int> cur;
        Helper(ret, cur, candidates, target, 0);
        return ret;
    }
    void Helper(vector<vector<int> > &ret, vector<int> cur, vector<int> &candidates, int target, int position){
        if(target == 0)
            ret.push_back(cur);
        else{
            for(int i = position; i < candidates.size() &&candidates[i] <= target; i ++){
                if(i != position && candidates[i] == candidates[i-1])
                    continue;
                cur.push_back(candidates[i]);
                Helper(ret, cur, candidates, target-candidates[i], i+1);
                cur.pop_back();
            }
        }
    }

};

int main(){
    int n; 
    cout<<"请输入向量长度："; 
    cin>>n;
    vector<int> nums;
    int a;
    for(int i=0;i<n;i++){
        cin>>a;
        nums.push_back(a);
    }
    int target;
    cout<<"请输入target:";
    cin>>target;
    Solution solution;
    vector<vector<int> > result=solution.combinationSum2(nums, target);
    int size = result.size();
    for(int i=0;i<size;i++){
        int size1=result[i].size();
        cout<<"[";
        int j;
        for(j=0;j<size1-1;j++){
            cout<<result[i][j]<<", ";   
        }
        cout<<result[i][j]<<"]"<<endl;
    }
}

No.3

题目链接：Unique Paths

  • 题目描述:

    A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

    The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

    How many possible unique paths are there?

这里写图片描述

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

(1)思路：动态规划，对于点(i， j)，只能从上边的点(i-1,j)或者左边的点(i,j-1)到达，且两者路径是不重复的，因此path[i][j] = path[i-1][j]+path[i][j-1]。

(2)代码：

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int> > path(m, vector<int>(n, 1));
        for(int i = 1; i < m; i ++) {
            for(int j = 1; j < n; j ++) {
                path[i][j] = path[i-1][j] + path[i][j-1];
            }
        }
        return path[m-1][n-1];
    }
};

(3)提交结果：

这里写图片描述

(4)测试代码：

#include<iostream>

#include<vector>

using namespace std;

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int> > path(m, vector<int>(n, 1));
        for(int i = 1; i < m; i ++)
        {
            for(int j = 1; j < n; j ++)
            {
                path[i][j] = path[i-1][j] + path[i][j-1];
            }
        }
        return path[m-1][n-1];
    }
};

int main(){
    int m, n;
    cout<<"请输入m n ：";
    cin>>m>>n;
    Solution solution;
    int result = solution.uniquePaths(m, n);
    cout<<result<<endl;
    return 0;
}

No.4

题目链接：Permutations

  • 题目描述：

    Given a collection of distinct numbers, return all possible permutations.

    For example,[1,2,3] have the following permutations:

    [
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

(1)思路：我们可以采用递归的方式。设置一个中间位置变量pos，让它初始化为0。用一层循环让该位置的数与其他所有位置的数交换位置，在该循环中进行递归调用，调用时将pos设为pos+1。直至pos的值等于整数向量的size-1即表示完成。

(2)测试代码：

#include<iostream>

#include<string>

#include<vector>

using namespace std;

class Solution {
public:
    vector<vector<int> > permute(vector<int>& nums) {
       vector<vector<int> > ret;
        Helper(ret, nums, 0);
        return ret;  
    }

    void Helper(vector<vector<int> >& ret, vector<int> num, int pos)
    {
        if(pos == num.size()-1)
            ret.push_back(num);
        else
        {
            for(int i = pos; i < num.size(); i ++)
            {
                swap(num[pos], num[i]);
                Helper(ret, num, pos+1);
                swap(num[pos], num[i]);
            }
        }
    }
};

int main(){
    int n; 
    cout<<"请输入向量长度："; 
    cin>>n;
    vector<int> nums;
    int a;
    for(int i=0;i<n;i++){
        cin>>a;
        nums.push_back(a);
    }
    Solution solution;
    vector<vector<int> > result=solution.permute(nums);
    int size = result.size();
    for(int i=0;i<size;i++){
        int size1=result[i].size();
        cout<<"[";
        int j;
        for(j=0;j<size1-1;j++){
            cout<<result[i][j]<<", ";   
        }
        cout<<result[i][j]<<"]"<<endl;
    }
}
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