五大常用算法--DP

Posted 2021-03-20 harry1989

概念

【geekforgeeks】

Dynamic Programming is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of subproblems, so that we do not have to re-compute them when needed later. This simple optimization reduces time complexities from exponential to polynomial. For example, if we write simple recursive solution for Fibonacci Numbers, we get exponential time complexity and if we optimize it by storing solutions of subproblems, time complexity reduces to linear.

动态规划是对于单纯的递归的一个主要优化方法。当我们看到递归解法对于同一个输入重复调用时，我们能通过动态规划方法重复优化。方法是简单存储子问题的结果，这样我们在后序用到它们的时候就不需要再重复计算了。这个简单的优化将时间复杂度从指数减少到了多项式。比如，如果我们写一下fibonacci数列的简单递归解法，我们得到了指数的时间复杂性，但是如果我们通过存储子问题进行优化，时间复杂性减少到了线性。

应用场景

通用解体思路

动态规划四要素：

1. 状态

2. 状态转移方程

3. 初始化

4. 结果

题目练习

Fibonacci数列 leetcode 509

最大连续子序列 leetcode 53

// 输入: [-2,1,-3,4,-1,2,1,-5,4],
// 输出: 6
// 解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。
class Solution {
  public int maxSubArray(int[] nums) {
    int n = nums.length, maxSum = nums[0];
    for(int i = 1; i < n; ++i) {
      if (nums[i - 1] > 0) nums[i] += nums[i - 1];
      maxSum = Math.max(nums[i], maxSum);
    }
    return maxSum;
  }
}

 

通配符匹配 leetcode 44

// 输入:
// s = "adceb"
// p = "*a*b"
// 输出: true
// 解释: 第一个 ‘*‘ 可以匹配空字符串, 第二个 ‘*‘ 可以匹配字符串 "dce".

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        boolean[][] f = new boolean[m + 1][n + 1];
        
        f[0][0] = true;
        for(int i = 1; i <= n; i++){
            f[0][i] = f[0][i - 1] && p.charAt(i - 1) == ‘*‘;
        }
        
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                if(s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == ‘?‘){
                    f[i][j] = f[i - 1][j - 1];
                }
                if(p.charAt(j - 1) == ‘*‘){
                    f[i][j] = f[i][j - 1] || f[i - 1][j];
                }
            }
        }
        return f[m][n];
    }
}

参考资料

1. https://www.geeksforgeeks.org/dynamic-programming/

2. https://leetcode-cn.com/problems/wildcard-matching/solution/dong-tai-gui-hua-si-yao-su-by-a380922457-4/

3. https://leetcode-cn.com/problems/maximum-subarray/solution/zui-da-zi-xu-he-by-leetcode/