LeetCode-64-Minimum Path Sum(动态规划)-Medium
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1. 题意理解:
对于MxN矩阵,求解从(0,0)点到(M-1,N-1)点的路径中,每个方格中的数值之和最小的路径。其中,每次只能从左向右,从上向下移动;
2. 题目分析:
1. 典型的动态规划题目;
2. 记录过程中的结果,避免重复计算,否则会超时;
3. 解题代码:
public class Solution { private int maxPath=0; private int[][] recordArray; private int minSubPathSum(int[][] grid, int r, int c){ if(r==1 && c==1){ return grid[0][0]; } int subPathSumLeft=maxPath; int subPathSumTop=maxPath; int rIndex=r-1; int cIndex=c-1; if(c-1>0){ if(recordArray[rIndex][cIndex-1]==-1){ recordArray[rIndex][cIndex-1]=minSubPathSum(grid, r, c-1); } subPathSumLeft=recordArray[rIndex][cIndex-1]; //System.out.println("subPathSumLeft= "+subPathSumLeft); } if(r-1>0){ if(recordArray[rIndex-1][cIndex]==-1){ recordArray[rIndex-1][cIndex]=minSubPathSum(grid, r-1, c); } subPathSumTop=recordArray[rIndex-1][cIndex]; //System.out.println("subPathSumTop= "+subPathSumTop); } int min=subPathSumLeft<subPathSumTop?subPathSumLeft:subPathSumTop; return min+grid[rIndex][cIndex]; } public int minPathSum(int[][] grid) { int row=grid.length; int col=grid[0].length; recordArray=new int[row][col]; for(int i=0;i<row;i++){ for(int j=0;j<col;j++){ maxPath+=grid[i][j]; recordArray[i][j]=-1; } } return minSubPathSum(grid, row, col); }}
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