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C/C++每日一练(20230422)

网友: 文章列表 2024-03-22 03:45:45

C/C++每日一练(20230422)

目录

1. 存在重复元素  🌟

2. 组合总和  🌟🌟

3. 给表达式添加运算符  🌟🌟🌟

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1. 存在重复元素

给定一个整数数组，判断是否存在重复元素。

如果存在一值在数组中出现至少两次，函数返回 true 。如果数组中每个元素都不相同，则返回 false 。

示例 1:

输入: [1,2,3,1]
输出: true

示例 2:

输入: [1,2,3,4]
输出: false

示例 3:

输入: [1,1,1,3,3,4,3,2,4,2]
输出: true

出处：

https://edu.csdn.net/practice/26235228

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:bool containsDuplicate(vector<int> &nums){if (nums.empty()){return false;}sort(nums.begin(), nums.begin() + nums.size());for (int i = 0; i < nums.size() - 1; i++){if (nums[i] == nums[i + 1]){return true;}}return false;}
};
int main()
{Solution s;vector<int> nums = {1,2,3,1};cout << (s.containsDuplicate(nums) ? "true" : "false") << endl;nums = {1,2,3,4};cout << (s.containsDuplicate(nums) ? "true" : "false") << endl;nums = {1,1,1,3,3,4,3,2,4,2};cout << (s.containsDuplicate(nums) ? "true" : "false") << endl;return 0;
}

输出：

true
false
true

2. 组合总和

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

  • 所有数字（包括 target）都是正整数。
  • 解集不能包含重复的组合。 

示例 1：

输入：candidates = [2,3,6,7], target = 7,
输出：[[7],[2,2,3]]

示例 2：

输入：candidates = [2,3,5], target = 8,
输出：[[2,2,2,2],[2,3,3],[3,5]]

提示：

  • 1 <= candidates.length <= 30
  • 1 <= candidates[i] <= 200
  • candidate 中的每个元素都是独一无二的。
  • 1 <= target <= 500

以下程序实现了这一功能，请你填补空白处内容：

···c++
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<vector<int>> combinationSum(vector<int> &candidates, int target)
    {
        vector<vector<int>> res;
        dfs(candidates, 0, target, res);
        return res;
    }
private:
    vector<int> stack;
    void dfs(vector<int> &candidates, int start, int target, vector<vector<int>> &res)
    {
        if (target < 0)
        {
            return;
        }
        else if (target == 0)
        {
            res.push_back(stack);
        }
        else
        {
            for (int i = start; i < candidates.size(); i++)
            {
                stack.push_back(candidates[i]);
                _____________________________;
                stack.pop_back();
            }
        }
    }
};
```

出处：

https://edu.csdn.net/practice/26235229

代码：

#include <bits/stdc++.h>
using namespace std;class Solution
{
public:vector<vector<int>> combinationSum(vector<int> &candidates, int target){vector<vector<int>> res;dfs(candidates, 0, target, res);return res;}
private:vector<int> stack;void dfs(vector<int> &candidates, int start, int target, vector<vector<int>> &res){if (target < 0){return;}else if (target == 0){res.push_back(stack);}else{for (int i = start; i < candidates.size(); i++){stack.push_back(candidates[i]);dfs(candidates, i, target - candidates[i], res);stack.pop_back();}}}
};string ArrayToString(vector<int> arr){string res = "[";int size = arr.size();for (int i = 0; i < size; i++) {res += to_string(arr[i]);if (i != size-1) {res += ",";}}return res + "]";
}void PrintArrays(vector<vector<int>> vect){cout << "[";int size = vect.size();for (int i = 0; i < size; i++) {cout << ArrayToString(vect[i]);if (i != size-1) {cout << ",";}}cout << "]" << endl;
}int main()
{Solution s;vector<int> candidates = {2,3,6,7};PrintArrays(s.combinationSum(candidates, 7));candidates = {2,3,5};PrintArrays(s.combinationSum(candidates, 8));return 0;
}

输出：

[[2,2,3],[7]]
[[2,2,2,2],[2,3,3],[3,5]]

3. 给表达式添加运算符

给定一个仅包含数字 0-9 的字符串 num 和一个目标值整数 target ，在 num 的数字之间添加 二元 运算符（不是一元）+、- 或 * ，返回所有能够得到目标值的表达式。

示例 1:

输入: num = "123", target = 6
输出: ["1+2+3", "1*2*3"]

示例 2:

输入: num = "232", target = 8
输出: ["2*3+2", "2+3*2"]

示例 3:

输入: num = "105", target = 5
输出: ["1*0+5","10-5"]

示例 4:

输入: num = "00", target = 0
输出: ["0+0", "0-0", "0*0"]

示例 5:

输入: num = "3456237490", target = 9191
输出: []

提示：

  • 1 <= num.length <= 10
  • num 仅含数字
  • -2^31 <= target <= 2^31 - 1

出处：

https://edu.csdn.net/practice/26235230

代码：

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:vector<string> addOperators(string num, int target){vector<string> res;addOperatorsDFS(num, target, 0, 0, "", res);return res;}void addOperatorsDFS(string num, int target, long long diff, long long curNum, string out, vector<string> &res){if (num.size() == 0 && curNum == target)res.push_back(out);for (int i = 1; i <= num.size(); ++i){string cur = num.substr(0, i);if (cur.size() > 1 && cur[0] == '0')return;string next = num.substr(i);if (out.size() > 0){addOperatorsDFS(next, target, stoll(cur), curNum + stoll(cur), out + "+" + cur, res);addOperatorsDFS(next, target, -stoll(cur), curNum - stoll(cur), out + "-" + cur, res);addOperatorsDFS(next, target, diff * stoll(cur), (curNum - diff) + diff * stoll(cur), out + "*" + cur, res);}elseaddOperatorsDFS(next, target, stoll(cur), stoll(cur), cur, res);}}
};string ArrayToString(vector<string> arr){string res = "[";int size = arr.size();for (int i = 0; i < size; i++) {res += arr[i];if (i != size-1) {res += ",";}}return res + "]";
}int main()
{Solution s;string num = "123";cout << ArrayToString(s.addOperators(num, 6)) << endl;num = "232";cout << ArrayToString(s.addOperators(num, 8)) << endl;num = "105";cout << ArrayToString(s.addOperators(num, 5)) << endl;num = "00";cout << ArrayToString(s.addOperators(num, 0)) << endl;num = "3456237490";cout << ArrayToString(s.addOperators(num, 9191)) << endl;num = "3236";cout << ArrayToString(s.addOperators(num, 24)) << endl;return 0;
}

输出：

[1+2+3,1*2*3]
[2+3*2,2*3+2]
[1*0+5,10-5]
[0+0,0-0,0*0]
[]
[3*2+3*6,3*2*3+6]

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