動態規劃小結 - 二維動態規劃 - 時間複雜度 O(n*n)的棋盤型，題 [LeetCode] Minimum Path Sum，Unique Paths II，Edit Distance

引言

二維動態規劃中最多見的是棋盤型二維動態規劃。

即

func(i, j) 每每只和 func(i-1, j-1), func(i-1, j) 以及 func(i, j-1) 有關

這種狀況下，時間複雜度 O(n*n)，空間複雜度每每能夠優化爲O(n)

 

例題  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.

時間複雜度 O(n*n)，空間複雜度O(n)的解法。

這裏用了個之前不用的技巧，當想把數組初始化爲非0的值時，不用memset，而改用vector表示數組。

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];
    }
};

 

 

例題  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.

 

時間複雜度 O(n*n)，空間複雜度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];
    }
};

 

 

例題  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

相似的還有http://basicalgos.blogspot.com/上的 53. Edit Distance between strings

這道題目咱們要設置兩個變量，i和j, E(i, j) 表示word1中以i 結尾的子串到表示word2中以j 結尾的子串的距離。

狀態轉移方程藉助53. Edit Distance between strings上的這張圖片來講明：

 

LeetCode那道題的實現代碼，時間複雜度 O(n*n)，空間複雜度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];
    }
};

 

 

後記

棋盤型二維動態規劃典型的題目還有「尋找最長公共子串(substring)」，「尋找最長公共子序列(subsequence)」。

這些均可以給出時間複雜度 O(n*n)，空間複雜度O(n)的解。