动态规划问题的三种类型

1.求最值

交换硬币 （lintcode 669）

public class CoinChange {

    public static void main(String[] args) {
        Solution solution = new CoinChange().new Solution();
        System.out.println("solution.coinChange(new int[]{1, 2, 5}, 11) = " + solution.coinChange(new int[]{1, 2, 5}, 11));
    }

    class Solution {
        public int coinChange(int[] A, int M) {
            int[] f = new int[M + 1];
            f[0] = 0;
            for (int i = 1; i <= M; i++) {
                f[i] = Integer.MAX_VALUE;
                for (int j = 0; j < A.length; j++) {
                    if (i >= A[j] && f[i - A[j]] != Integer.MAX_VALUE) {
                        f[i] = Math.min(f[i - A[j]] + 1, f[i]);
                    }
                }
                System.out.printf("f[%s] = %s\n", i, f[i]);
            }
            if (f[M] == Integer.MAX_VALUE) {
                f[M] = -1;
            }
            return f[M];
        }
    }
}

 

2.求总数

机器人路径（lintcode 114）

public class UniquePath {

    public static void main(String[] args) {
        Solution solution = new UniquePath().new Solution();
        System.out.println(solution.uniquePath(1, 1));
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int uniquePath(int m, int n) {
            int[][] dp = new int[m][n];
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                    }
                }
            }
            return dp[m - 1][n - 1];
        }
    }
//leetcode submit region end(Prohibit modification and deletion)
}

　　

3.存在型

跳跃游戏（lintcode 116）

public class JumpGame {

    public static void main(String[] args) {
        Solution solution = new JumpGame().new Solution();

    }

    class Solution {
        public boolean canJump(int[] A) {
            int n = A.length;
            boolean[] f = new boolean[n];
            f[0] = true;
            for (int j = 1; j < n; j++) {
                f[j] = false;
                for (int i = 0; i < j; i++) {
                    if (f[i] && i + A[i] >= j) {
                        f[j] = true;
                        break;
                    }
                }
            }
            return f[n - 1];
        }
    }
}

　　

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