算法系列博客之Dynamic Programming
本篇博客將運用動態規劃的思想來解決leetcode上264號問題
這三個題目的共同之處在於均是二維矩陣上的規劃問題
問題描述：
62 Unique Paths
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).
How many possible unique paths are there?
由於只能向下或者向右走一步，因而其狀態僅僅依賴於左側和上側，轉化爲數字形式即
假定paths[i][j]表示從(0,0)到(i,j)的總unique paths數目，則：
paths[i][j] = paths[i-1][j] + paths[i][j-1]
觀察發現，其狀態依賴也僅僅只依賴於上一行和本行前列狀態，因而可以只存儲兩行狀態
特別地，paths[i][0] = paths[0][j] = 1，代碼實現如下
class Solution(object):
def uniquePaths(self, m, n):
paths = [[1] * n, [0] * n]
for i in range(1, m):
paths[1][0] = 1
for j in range(1, n):
paths[1][j] = paths[1][j-1] + paths[0][j]
paths[0] = paths[1]
return paths[0][n-1]時間複雜度分析，兩層無嵌套常數條指令的循環，爲O(m*n)
空間複雜度分析，僅存儲兩行狀態，每行n列，爲O(n)
63 Unique Paths II
Follow up for “Unique Paths”:
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.
此題較上題不同的是，某些點可能無法通過
但仍然是隻能向右或者向下走一步，因而狀態依賴關係依然沒有改變，僅需要添加額外約束
即如果點(i,j)不能通過，則paths[i][j] = 0
如果點(i,j)能通過，則paths[i][j] = paths[i-1][j] + paths[i][j-1]
且初始化狀態只能有一個，即paths[0][0] = 1或0
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
if obstacleGrid[0][0] == 1: return 0
paths = [[1] * len(obstacleGrid[0])] * 2
for j in range(1, len(obstacleGrid[0])):
paths[0][j] = 0 if obstacleGrid[0][j] else paths[0][j-1]
for i in range(1, len(obstacleGrid)):
paths[1][0] = 0 if obstacleGrid[i][0] else paths[0][0]
for j in range(1, len(obstacleGrid[0])):
paths[1][j] = 0 if obstacleGrid[i][j] else paths[1][j-1] + paths[0][j]
paths[0] = paths[1]
return paths[0][len(obstacleGrid[0])-1]複雜度較之前沒有改變，仍然是時間O(m*n)，空間O(n)
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.
題目的模型較之前不同的是，每個點都增加了權值grid[i][j]
而目標是要找出權值最小的路徑，但是仍然有隻能向右或者向下的約束
因此仍然可以採用上面的那種策略，僅僅是狀態轉移方程稍微不一樣
假設minpath[i][j]表示從(0,0)到(i,j)的最小權值路徑，則
minpath[i][j] = grid[i][j] + min(minpath[i][j-1], minpath[i-1][j])
特別地，minpath[0][0] = grid[0][0];
minpath[0][j] = grid[0][j] + minpath[0][j-1];
minpath[i][0] = grid[i][0] + minpath[i-1][0]
python代碼實現如下
class Solution(object):
def minPathSum(self, grid):
minpath = [[x for x in grid[0]], [0]*len(grid[0])]
for j in range(1, len(grid[0])):
minpath[0][j] += minpath[0][j-1]
for i in range(1, len(grid)):
minpath[1][0] = grid[i][0] + minpath[0][0]
for j in range(1, len(grid[0])):
minpath[1][j] = grid[i][j] + min(minpath[1][j-1], minpath[0][j])
minpath[0] = minpath[1]
return minpath[0][len(grid[0])-1]同樣，複雜度並沒有改變，時間O(m*n)，空間O(n)