[IM]: leetcode

03.code/08.leetcode/03.7.8刷题/00.0-1背包问题.cpp

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+#include <iostream>
+#include <vector>
+#include <cassert>
+using namespace std;
+
+// 有一个背包，他的容量为C(Capacity)。现在又n种不同的物品，编号为0....n - 1 ,其中每一件物品的重量为w(i),价值为v(i).问可以向这个背包中盛放哪些物品，使得在不超过背包容量的基础上，物品的总价值最大
+// 暴力解法：每一件物品都可以放进背包，也可以不放进去
+// 动态规划：
+// 状态定义：
+// F(n,C) ：将n个物品放进容量为C的背包，使得价值最大
+// F(i, c) = F(i - 1, c)                           // i不放入背包
+//         = v(i) + F(i - 1, c- W(i))             // i放入背包
+class Solution
+{
+private:
+	// 用[0 ... index]的物品，填充容积为C的背包的最大价值
+	int bestValue(const vector<int> &w, const vector<int> v, int index, int c)
+	{
+		if (index < 0 || c <= 0)
+		{
+			return 0;
+		}
+
+		int res = bestValue(w, v, index - 1, c);
+		if (c >= w[index])
+		{
+			res = max(res, v[index] + bestValue(w, v, index - 1, c - w[index]));
+		}
+
+		return res;
+	}
+
+	// 动态规划 + 记忆化搜索
+	vector<vector<int>> memo; // 
+	int bestValue2(const vector<int> &w, const vector<int> v, int index, int c)
+	{
+		if (index < 0 || c <= 0)
+		{
+			return 0;
+		}
+
+		if (memo[index][c] != -1)
+		{
+			return memo[index][c];
+		}
+
+		int res = bestValue(w, v, index - 1, c);
+		if (c >= w[index])
+		{
+			res = max(res, v[index] + bestValue(w, v, index - 1, c - w[index]));
+		}
+
+		memo[index][c] = res;
+
+		return res;
+	}
+public:
+	int knapspack01(const vector<int> &w, const vector<int> &v, int c)
+	{
+		int n = w.size();
+		return bestValue(w, v, n - 1, c);
+	}
+	int knapspack02(const vector<int> &w, const vector<int> &v, int c)
+	{
+		int n = w.size();
+
+		memo = vector<vector<int>> (n, vector<int>(c + 1, -1)); // n * C + 1
+		return bestValue2(w, v, n - 1, c);
+	}
+	int knapspack03(const vector<int> &w, const vector<int> &v, int c)
+	{
+		int n = w.size();
+		if (0 == n)
+		{
+			return 0;
+		}
+		vector<vector<int>> memo(n, vector<int>(c + 1, -1));
+
+		// 先考虑0号物品
+		for (int j = 0; j <= c; ++j)
+		{
+			memo[0][j] = ( j >= w[0] ? v[0] : 0);
+		}
+
+		for (int i = 1; i < n; ++i)
+		{
+			for (int j = 0; j <= c; ++j)
+			{
+				memo[i][j] = memo[i - 1][j];
+				if (j >= w[i])
+				{
+					memo[i][j] = max( memo[i][j], v[i] + memo[i - 1][j - w[i]]);
+				}
+			}
+		}
+
+		return memo[n - 1][c];
+	}
+
+	int knapspack04(const vector<int> &w, const vector<int> &v, int c)
+	{
+		int n = w.size();
+		if (0 == n)
+		{
+			return 0;
+		}
+		vector<vector<int>> memo(2, vector<int>(c + 1, -1));
+
+		// 先考虑0号物品
+		for (int j = 0; j <= c; ++j)
+		{
+			memo[0][j] = ( j >= w[0] ? v[0] : 0);
+		}
+
+		for (int i = 1; i < n; ++i)
+		{
+			for (int j = 0; j <= c; ++j)
+			{
+				memo[i % 2][j] = memo[(i - 1) % 2][j];
+				if (j >= w[i])
+				{
+					memo[i % 2][j] = max( memo[i % 2][j], v[i] + memo[(i - 1) % 2][j - w[i]]);
+				}
+			}
+		}
+
+		return memo[(n - 1) % 2][c];
+	}
+
+};
+
+
+int main() {
+
+	vector<int> w = {1, 2, 3};
+	vector<int> v = {6, 10, 12};
+	int c = 5;
+
+	Solution sol;
+	cout << sol.knapspack01(w, v, c) << endl;
+	cout << sol.knapspack02(w, v, c) << endl;
+	cout << sol.knapspack03(w, v, c) << endl;
+	cout << sol.knapspack04(w, v, c) << endl;
+
+
+	return 0;
+}

03.code/08.leetcode/03.7.8刷题/120.cpp

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+#include <iostream>
+#include <vector>
+using namespace std;
+
+class Solution {
+public:
+    int min(vector<vector<int>> &triangle)
+    {
+        size_t size = triangle.size();
+        for (int i = triangle.size(); i > 0; --i)
+        {
+            for (int j = 0; j < triangle[size].size(); ++j)
+            {
+                triangle[size][j] = min(triangle[i][j - 1], triangle[i][j]);
+            }
+        }
+
+        return 0;
+    }
+    int minimumTotal(vector<vector<int>>& triangle) {
+
+    }
+};
+
+int main()
+{
+
+	return 0;
+}

03.code/08.leetcode/03.7.8刷题/343.cpp

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+#include <iostream>
+#include <vector>
+using namespace std;
+
+int integerBreak(int n) {
+	vector<int> dp(n + 1, 0);
+	dp[1] = 1;
+
+	for (int i = 0; i < n + 1; ++i)
+	{
+		for (int j = i - 1; j >= 1; --j)
+		{
+			dp[i] = max(dp[i], dp[j] * (i - j));
+			cout << "max1:" << dp[i] << ", " << dp[j] * (i - j) << endl;
+			dp[i] = max(dp[i], j * (i - j));
+			cout << "max2:" << dp[i] << ", " << j * (i - j) << endl;
+		}
+		cout << endl;
+	}
+
+	return dp[n];
+}
+
+/**
+     * 1. 确定dp数组下标含义 分拆数字i，可以得到的最大乘积为dp[i];
+     * 2. 递推公式 dp[i] = max(dp[i], (i - j) * j, dp[i - j] * j);
+     * 3. 初始化 dp[2] = 1;
+     * 4. 遍历顺序 从前向后遍历就可以;
+     * 5. 推导结果;
+     */
+int integerBreak2(int n)
+{
+	/* 定义dp数组 */
+	vector<int> dp(n + 1, 0);
+	/* dp数组初始化 */
+	dp[2] = 1;
+	/* 从前向后遍历 */
+	cout << "dp[i] " << ",dp[i - j] * j" << ",((i - j) * j)" << endl;
+	for (int i = 3; i <= n ; i++)
+	{
+		/* j遍历到小于i的值 */
+		for (int j = 1; j < i - 1; j++)
+		{
+			cout << dp[i] << " ";
+			dp[i] = max(dp[i], dp[i - j] * j);
+			cout << dp[i - j] * j << " ";
+			dp[i] = max(dp[i], ((i - j) * j));
+			cout << ((i - j) * j) << " | ";
+		}
+		cout << "dp[" << i << "]->" << dp[i] << endl;
+	}
+	return dp[n];
+}
+
+int integerBreak3(int n)
+{
+	if (2 == n)
+	{
+		return 1;
+	}
+
+	return 0;
+}
+
+int main() {
+
+	cout << integerBreak2(10) << endl;
+	return 0;
+}

03.code/08.leetcode/03.7.8刷题/37.cpp

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+#include <iostream>
+#include <cmath>
+#include <cstdio>
+#include <cstdlib>
+#include <vector>
+#include <stack>
+#include <queue>
+#include <string>
+#include <map>
+#include <set>
+#include <algorithm>
+using namespace std;
+
+vector<vector<bool>> col = vector<vector<bool>>(9, vector<bool>(9, false));
+
+void print(vector<vector<char>>& v)//, vector<vector<bool>> &col)
+{
+	for_each(v.begin(), v.end(),
+	[&](vector<char> &w){
+		for_each(w.begin(), w.end(), [&](char &val){ cout << val << ' '; });
+		cout << endl;
+	});
+
+	// for_each(col.begin(), col.end(),
+	// [&](vector<bool> &w){
+	// 	for_each(w.begin(), w.end(), [&](bool val){ cout << val << ' '; });
+	// 	cout << endl;
+	// });
+}
+
+int c = 0;
+
+vector<vector<char>> ans;
+
+void dfs(int x, int y, vector<char> &v, vector<char> &row)
+{
+	if (y == v.size())
+	{
+		ans.push_back(row);
+		return ;
+	}
+
+	row.push_back(v[y]);
+	if ('.' == row[y])
+	{
+		char tmp;
+		for (int j = 0; j < col[x].size(); ++j)
+		{
+			if (!col[x][j])
+			{
+				tmp = j + '0';
+			}
+		}
+		col[x][tmp - '0'] = true;
+		row[y] = tmp;
+	}
+	dfs(x, y + 1, v, row);
+
+}
+
+void solveSudoku(vector<vector<char>>& v)
+{
+	for_each(v.begin(), v.end(),
+		[&](vector<char> &w){
+			for_each(w.begin(), w.end(), 
+			[&](char &val){ 
+				if ('.' != val)
+				{
+					col[c][val - '0'] = true;
+				}
+			 });
+			++c;
+		});
+
+
+	print(v);//, col);
+}
+
+int main()
+{
+	vector<vector<char>> v = {
+		{'5','3','.','.','7','.','.','.','.'},{'6','.','.','1','9','5','.','.','.'},{'.','9','8','.','.','.','.','6','.'},
+		{'8','.','.','.','6','.','.','.','3'},{'4','.','.','8','.','3','.','.','1'},{'7','.','.','.','2','.','.','.','6'},
+		{'.','6','.','.','.','.','2','8','.'},{'.','.','.','4','1','9','.','.','5'},{'.','.','.','.','8','.','.','7','9'}
+		};
+
+	solveSudoku(v);
+
+	// char c = 2 + '0';
+	// cout << c << endl;
+	vector<char> v1 = {'5','3','.','.','7','.','.','.','.'};
+	vector<char> row;
+	dfs(0, 0, v1, row);
+	cout << "----" << endl;
+	print(ans);
+	// for_each(v1.begin(), v1.end(), [](char &v) { cout << v << " ";});
+
+
+	return 0;
+}