动态规划

509 斐波那契数列

class Solution {
    public int fib(int n) {
        int[] dp = new int[n+1];
        if (n <= 1) {
            return n;
        }
        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];
        if (n <= 1) {
            return n;
        }
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; 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 n = cost.length;
        int[] dp = new int[n];
        dp[0] = cost[0];
        dp[1] = cost[1];
        for (int i = 2; i < n; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return Math.min(dp[n - 1], dp[n - 2]);
    }
}

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 ll

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        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++) {
                if (obstacleGrid[i][j] == 0) {
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
        }
        return dp[m-1][n-1];
    }
}

343 整数拆分

//分拆数字i，可以得到的最大乘积为dp[i]
//dp[i] = max(dp[i], max(j*(i-j), j*dp[i-j]))
//(i-j)拆分+不拆分取最大
class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n+1];
        dp[2] = 1;
        for (int i = 3; i <= n; i++) {
            for (int j = 0; j < i - 1; j++) {
                dp[i] = Math.max(dp[i], Math.max(j * (i - j), j * dp[i - j]));
            }
        }
        return dp[n];
    }
}

96 不同的二叉搜索树

//dp[i]: 1到i为节点组成的二叉搜索树的个数为dp[i]
class Solution {
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= i; j++) {
                dp[i] += dp[j - 1] * dp[i - j];
            }
        }
        return dp[n];
    }
}