动态规划专题

Posted 2020-11-23 yangcao

一、记忆化搜索

 斐波那契算法

1.递归代码（时间复杂度O(2^n)）：

int f(int n){  
    if(n == 1 || n == 2){  
        return 1;  
    }  
    return f(n-1) + f(n-2);  
} 

2.递归加记忆化

public class Solution{

    int[] res = new int[n];
    Arrays.fill(res, -1);

    int f(int n){  
    if(n == 1 || n == 2){  
        return 1;  
    }  
    if(res[n] == -1) return f(n-1) + f(n-2);  

    return res[n]; 
   } 
}
    

 

3.动态规划  或者 转化为非递归 或者转化为矩阵求解

 

递归是自上而下的解决问题（如求f(n)就已经假设所有小于n的值已经得到，）

动态规划是自下而上的解决问题（求f(n) 就已经把前面n-1个值完全获得）

动态规划   约等于  记忆化搜索 （严格来说不对 因而 记忆化搜索也是自上而下，动态规划是自下而上的解决问题 ）

 

70. Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

代码如下：

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

 

120. Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

代码如下：

class Solution {
  public int minimumTotal(List<List<Integer>> triangle) {
      int[] re = new int[triangle.size() + 1];
      for(int i=triangle.size()-1;i>=0; i--){
          for(int j=0; j<triangle.get(i).size(); j++){
              re[j] = Math.min(re[j],re[j+1]) + triangle.get(i).get(j);//re数组每次填充的值都是一样的
          }
      }
      return re[0];
  }
}

 

64. Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

代码如下：

class Solution {
    public int minPathSum(int[][] grid) {
        
        for(int i=1; i<grid.length; i++) {
            grid[i][0] = grid[i-1][0] + grid[i][0];
        }
        for(int j=1; j<grid[0].length; j++) {
            grid[0][j] = grid[0][j-1] + grid[0][j];
        }
        for(int i=1; i<grid.length; i++) {
            for(int j=1; j<grid[0].length; j++) {
                grid[i][j] = Math.min(grid[i-1][j], grid[i][j-1]) + grid[i][j];
            }
        }
        return grid[grid.length-1][grid[0].length-1];
    }
}

 

343. Integer Break

Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.
代码如下：