代码随想录-动态规划

对于动态规划问题，我将拆解为如下五步曲，这五步都搞清楚了，才能说把动态规划真的掌握了！

1. 确定dp数组（dp table）以及下标的含义
2. 确定递推公式
3. dp数组如何初始化
4. 确定遍历顺序
5. 举例推导dp数组

509.斐波那契数

class Solution {
   
    public int fib(int n) {
   
        if (n <= 1) {
   
            return n;
        }
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
   
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

70. 爬楼梯

class Solution {
   
    public int climbStairs(int n) {
   
        int[] dp = new int[n+1];
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2;i <= n; i++){
   
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
}

746.使用最小花费爬楼梯

class Solution {
   
    public int minCostClimbingStairs(int[] cost) {
   
        int len = cost.length;
        int[] dp = new int[len + 1];

        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i <= len; i++) {
   
            dp[i] = Math.min((dp[i - 1] + cost[i - 1]), (dp[i - 2] + cost[i - 2]));
        }
        return dp[len];
    }
}

62. 不同路径

class Solution {
   
    public int uniquePaths(int m, int n) {
   
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
   
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
   
            dp[0][i] = 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];
    }
}

63. 不同路径 II

class Solution {
   
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
   
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];

        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
   
            return 0;
        }

        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
   
            dp[i][0] = 1;
        }
        for (int i = 0; i < n && obstacleGrid[0][i] == 0; i++) {
   
            dp[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
   
            for (int j = 1; j < n; j++) {
   
                dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;
            }
        }
        return dp[m - 1][n - 1];
    }
}

343. 整数拆分

class Solution {
   
    public int integerBreak(int n) {
   
        int[] dp = new int[n + 1];
        for (int i = 2; i <= n; i++) {
   
            int curMax = 0;
            for (int j = 1; j < i; j++) {
   
                curMax = Math.max(curMax, Math.max(j * (i - j), j * dp[i - j]));
            }
            dp[i] = curMax;
        }
        return dp[n];
    }
}

96. 不同的二叉搜索树

class Solution {
   
    public int numTrees(int n) {
   
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
   
            for (int j = 1; j <= i; j++) {
   
                dp[i] += dp[j - 1] * dp[i - j];
            }
        }
        return dp[n];
    }
}

01背包和完全背包就够用了。
而完全背包又是也是01背包稍作变化而来，即：完全背包的物品数量是无限的。

01背包

01背包

// weight数组的大小 就是物品个数
for(int i = 1; i < weight.size(); i++) {
    // 遍历物品
    for(int j = 0; j <= bagweight; j++) {
    // 遍历背包容量
        if (j < weight[i]) dp[i][j] = dp[i - 1][j];
        else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);

    }
}

public class BagProblem {
   
    public static void main(String[] args) {
   
        int[] weight = {
   1,3,4};
        int[] value = {
   15,20,30};
        int bagSize = 4;
        testWeightBagProblem(weight,value,bagSize);
    }

    /**
     * 动态规划获得结果
     * @param weight  物品的重量
     * @param value   物品的价值
     * @param bagSize 背包的容量
     */
    public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
   

        // 创建dp数组
        int goods = weight.length;  // 获取物品的数量
        int[][] dp = new int[goods][bagSize + 1];

        // 初始化dp数组
        // 创建数组后，其中默认的值就是0