动态规划小结

引言

二维动态规划中最常见的是棋盘型二维动态规划。

即

func(i, j) 往往只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有关

这种情况下，时间复杂度 O(n*n)，空间复杂度往往可以优化为O(n)

例题  1

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.

时间复杂度 O(n*n)，空间复杂度O(n)的解法。

这里用了个以前不用的技巧，当想把数组初始化为非0的值时，不用memset，而改用vector表示数组。

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        if(grid.size() == 0 || grid[0].size() == 0) return 0;
        int H = grid.size(), W = grid[0].size();
        vector<int> path(W+1, INT_MAX);
        path[1] = 0;
        for(int i = 1; i <= H; ++i)
            for(int j = 1; j <= W; path[j] = min(path[j-1], path[j]) + grid[i-1][j-1], ++j);
        return path[W];
    }
};

例题  2

Unique Paths II

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

Above is a 3 x 7 grid. How many possible unique paths are there?

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

时间复杂度 O(n*n)，空间复杂度O(n)

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        if(obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0) return 0;
        int H = obstacleGrid.size(), W = obstacleGrid[0].size();
        int paths[W+1]; memset(paths, 0, sizeof(paths));
        paths[1] = (obstacleGrid[0][0] ? 0 : 1);
        for(int i = 1; i <= H; ++i){
            for(int j = 1; j <= W; ++j){
                paths[j] = (obstacleGrid[i-1][j-1] ? 0 : paths[j-1] + paths[j]);
            }
        }
        return paths[W];
    }
};

例题  3

很熟悉的 Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

类似的还有http://basicalgos.blogspot.com/上的 53. Edit Distance between strings

这道题目我们要设置两个变量，i和j, E(i, j) 表示word1中以i 结尾的子串到表示word2中以j 结尾的子串的距离。

状态转移方程借助53. Edit Distance between strings上的这张图片来说明：

LeetCode那道题的实现代码，时间复杂度 O(n*n)，空间复杂度O(n)

class Solution {
public:
    int minDistance(string word1, string word2) {
        if(word1.empty() && word2.empty()) return 0;
        int len1 = word1.length(), len2 = word2.length();
        int A[len2+1]; int pre;
        memset(A, 0, sizeof(A));
        for(int i = 0; i <= len1; ++i){
            for(int j = 0; j <= len2; ++j){
                int Min = INT_MAX;
                if(i > 0) Min = min(A[j]+1, Min);
                if(j > 0) Min = min(A[j-1]+1, Min);
                if(i > 0 && j > 0) Min = min(Min, pre+(word1[i-1] == word2[j-1] ? 0 : 1));
                if(i == 0 && j == 0) Min = 0;
                pre = A[j];
                A[j] = Min;
            }
        }
        return A[len2];
    }
};

后记

棋盘型二维动态规划典型的题目还有“寻找最长公共子串(substring)”，“寻找最长公共子序列(subsequence)”。

这些都可以给出时间复杂度 O(n*n)，空间复杂度O(n)的解。