update from laptop0520

Author: codingClaire

.vscode/launch.json

@@ -6,7 +6,7 @@
     "configurations": [
 
         {
-            "name": "gcc.exe - 生成和调试活动文件",
+            "name": "gcc.exe - debug",
             "type": "cppdbg",
             "request": "launch",
             "program": "${fileDirname}\\${fileBasenameNoExtension}.exe",

.vscode/tasks.json

@@ -21,7 +21,11 @@
                 "kind": "build",
                 "isDefault": true
             },
-            "detail": "调试器生成的任务。"
+            "detail": "调试器生成的任务。",
+            //从这里开始是自己添加的
+            "presentation": { 
+                "panel": "new" //默认为“shared“表示共享，改成new之后每个进程创建新的端口
+            }//添加这一整块代码块
         }
     ],
     "version": "2.0.0"

codingInterview/58II.左旋转字符串.cpp renamed to codingInterview/58 - II. 左旋转字符串.cpp

@@ -14,13 +14,17 @@ class Solution
     // 9个数选择k个数，让它们的和为n
     void DFS(int cur, int n,int k, int sum)
     {
+        //当前选择的数 加上 剩余的个数也到不了k 
+        //那么肯定全部加上也不够k个，直接剪枝
         if(temp.size()+(n-cur+1)<k || temp.size()>k)
             return;
+        //保存条件： tmp存了k个数 并且加起来要等于n的情况
         if(temp.size() == k  && accumulate(temp.begin(),temp.end(),0) == sum)
         {
             ans.emplace_back(temp);
             return;
         }
+        //每个数，选或不选
         temp.emplace_back(cur);
         DFS(cur+1, n,k,sum);
         temp.pop_back();

totalProblem(CN)/216.组合总和-iii.cpp

@@ -7,6 +7,8 @@
 // @lc code=start
 //[34] 在排序数组中查找元素的第一个和最后一个位置
 // 92.64 % 16.44 %
+# include<bits/stdc++.h>
+using namespace std;
 class Solution
 {
 public:
@@ -131,7 +133,7 @@ class Solution2
 int main()
 {
     Solution2 A;
-    vector<int> nums{2, 2};
+    vector<int> nums{2, 2, 3};
     int target = 3;
     vector<int> res;
     res = A.searchRange(nums, target);

totalProblem(CN)/34.在排序数组中查找元素的第一个和最后一个位置.cpp

@@ -21,9 +21,10 @@ class Solution
             res.emplace_back(combine);
             return;
         }
-        // 选择不用，直接跳过
+        // 选择不用，直接跳过   
+        // 不选，也要有一个尽头
         DFS(candidates, target, res, combine, idx + 1);
-        // 选择当前数
+        // 选择当前数, 并且idx不会往下更新
         if (target - candidates[idx] >= 0)
         {
             combine.emplace_back(candidates[idx]);
@@ -41,4 +42,37 @@ class Solution
     }
 };
 // @lc code=end
+// 二刷
+class Solution {
+public:
 
+    void dfs(vector<vector<int>>&ans , vector<int>& tmp, vector<int>& candidates, int target, int idx)
+    {
+        if(idx == candidates.size())
+            return;
+        if(target == 0)
+        {
+            ans.emplace_back(tmp);
+            return;
+        }
+        // 要注意的是这里要加1
+        dfs(ans, tmp, candidates,target,idx+1);
+        // 这一句判断，很重要！！
+        if(target-candidates[idx]>=0)
+        {
+            tmp.emplace_back(candidates[idx]);
+            dfs(ans, tmp, candidates,target-candidates[idx],idx);
+            tmp.pop_back();
+        }
+        return;
+    }
+
+
+    vector<vector<int>> combinationSum(vector<int>& candidates, int target) 
+    {
+        vector<vector<int>>  ans;
+        vector<int> tmp;
+        dfs(ans, tmp, candidates, target, 0);
+        return ans;
+    }
+};

totalProblem(CN)/39.组合总和.cpp

@@ -5,11 +5,69 @@
  */
 
 // @lc code=start
-class Solution {
+class Solution
+{
 public:
-    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
-
+    void dfs(vector<vector<int>> &ans, vector<int> &tmp, vector<int> &candidates, int target, int idx)
+    {
+        if (target == 0)
+        {
+            ans.emplace_back(tmp);
+            return;
+        }
+        for (int i = idx; i < candidates.size() && target - candidates[i] >= 0; i++)
+        {
+            if (i > idx && candidates[i] == candidates[i - 1])
+                continue;
+            tmp.emplace_back(candidates[i]);
+            dfs(ans, tmp, candidates, target - candidates[i], i + 1);
+            tmp.pop_back();
+        }
+    }
+    vector<vector<int>> combinationSum2(vector<int> &candidates, int target)
+    {
+        vector<vector<int>> ans;
+        vector<int> tmp;
+        sort(candidates.begin(), candidates.end());
+        dfs(ans, tmp, candidates, target, 0);
+        return ans;
     }
 };
 // @lc code=end
 
+// 以下的解法是没办法正确的
+class Solution
+{
+public:
+    void dfs(vector<vector<int>> &ans, vector<int> &tmp, vector<int> &candidates, int target, int idx)
+    {
+        if (idx == candidates.size())
+            return;
+        if (target == 0)
+        {
+            ans.emplace_back(tmp);
+            return;
+        }
+        dfs(ans, tmp, candidates, target, idx + 1);
+        //为什么呢？ 因为没办法确定前面是否选择了candidate[idx]
+        // 这样实际上就会导致漏掉一些结果！！ 
+        if (target - candidates[idx] >= 0 && candidates[idx] != candidates[idx + 1])
+        {
+            tmp.emplace_back(candidates[idx]);
+            dfs(ans, tmp, candidates, target - candidates[idx], idx + 1);
+            tmp.pop_back();
+        }
+        return;
+    }
+    vector<vector<int>> combinationSum2(vector<int> &candidates, int target)
+    {
+        vector<vector<int>> ans;
+        vector<int> tmp;
+        sort(candidates.begin(), candidates.end());
+        dfs(ans, tmp, candidates, target, 0);
+        return ans;
+    }
+};
+
+// 在示例里 第一个组合不会出现（[ 1, 1, 6 ]）
+// [[ 1, 1, 6 ], [ 1, 2, 5 ], [ 1, 7 ], [ 2, 6 ] ]

totalProblem(CN)/40.组合总和-ii.cpp

@@ -0,0 +1,29 @@
+/*
+ * @lc app=leetcode.cn id=509 lang=cpp
+ *
+ * [509] 斐波那契数
+ */
+
+// @lc code=start
+// 100 %  23.5 %
+class Solution 
+{
+public:
+    int fib(int n) 
+    {
+        vector<int> dp(n+1,0);
+        dp[0] = 0;
+        if(n == 0)
+            return dp[0];
+        dp[1] = 1;
+        if(n == 1)
+            return dp[1];
+        for(int i=2;i<=n;i++)
+        {
+            dp[i] = dp[i-1] + dp[i-2];
+        }
+        return dp[n];
+    }
+};
+// @lc code=end
+

totalProblem(CN)/509.斐波那契数.cpp

@@ -28,3 +28,28 @@ class Solution
     }
 };
 // @lc code=end
+// 二刷
+class Solution 
+{
+public:
+    int uniquePaths(int m, int n) 
+    {
+        vector<vector<int>> dp(m,vector<int>(n,0));
+        for(int i =0;i<m;i++)
+        {
+            dp[i][0] =1;
+        } 
+        for(int j=0;j<n;j++)
+        {
+            dp[0][j] = 1;
+        }
+        for(int i=1;i<m;i++)
+        {
+            for(int j=1;j<n;j++)
+            {
+                dp[i][j] = dp[i-1][j] +dp[i][j-1];
+            }
+        }
+        return dp[m-1][n-1];
+    }
+};

totalProblem(CN)/62.不同路径.cpp

@@ -0,0 +1,37 @@
+/*
+ * @lc app=leetcode.cn id=63 lang=cpp
+ *
+ * [63] 不同路径 II
+ */
+
+// @lc code=start
+
+class Solution
+{
+public:
+    int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid)
+    {
+        int m = obstacleGrid.size();
+        int n = obstacleGrid[0].size();
+        // 如果在起点或终点出现了障碍，直接返回0
+        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) 
+            return 0;
+        vector<vector<int>> dp(m, vector<int>(n, 0));
+        // 这里注意也需要加上判断条件
+        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++)
+            dp[i][0] = 1;
+        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++)
+            dp[0][j] = 1;
+        for (int i = 1; i < m; i++)
+        {
+            for (int j = 1; j < n; j++)
+            {
+                if (obstacleGrid[i][j] == 1)
+                    continue;
+                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
+            }
+        }
+        return dp[m - 1][n - 1];
+    }
+};
+// @lc code=end