标签：== rom nbsp cost div class simple get down
Given a 2 dimensional matrix, find minimum cost path to reach bottom right from top left provided you can only from down and right.
For a 5 * 5 matrix, think of solving this problem by breaking it into smaller subproblems. And overlapping subproblems is clear.
(4, 4)
(4, 3) (3,4)
(4, 2) (3,3) (3,3) (2,4)
1 public int getMinPathCost(int[][] matrix) { 2 if(matrix == null || matrix.length == 0 || matrix[0].length == 0) { 3 return 0; 4 } 5 int[][] cost = new int[matrix.length][matrix[0].length]; 6 cost[0][0] = matrix[0][0]; 7 for(int i = 1; i < cost.length; i++) { 8 cost[i][0] = cost[i - 1][0] + matrix[i][0]; 9 } 10 for(int j = 1; j < cost[0].length; j++) { 11 cost[0][j] = cost[0][j - 1] + matrix[0][j]; 12 } 13 for(int i = 1; i < cost.length; i++) { 14 for(int j = 1; j < cost[0].length; j++) { 15 cost[i][j] = Math.min(cost[i][j - 1], cost[i - 1][j]) + matrix[i][j]; 16 } 17 } 18 return cost[cost.length - 1][cost[0].length - 1]; 19 }
[Coding Made Simple] Minimum Cost Path
标签：== rom nbsp cost div class simple get down
原文地址：http://www.cnblogs.com/lz87/p/7288839.html
