标签：style blog http io ar color sp for on
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
用DP主要还是要找出递推关系，边界条件
这里的递推关系miniPaht[i][j]表示第i层第j个点的最短路径
miniPath[i][j] = min{miniPath[i - 1][j], miniPath[i - 1][j - 1]} + triangle.get(i).get(j)
注意i和j等于0和等于 triangle.get(i).size - 1的边界条件问题
求出最后一层所有点的最小值，在遍历最后一层找出最小值
参考：http://blog.csdn.net/zhanghaodx082/article/details/24599479
1 public class Solution { 2 public int minimumTotal(List<List<Integer>> triangle) { 3 if(null == triangle) 4 return 0; 5 if(1 == triangle.size()) 6 return triangle.get(0).get(0); 7 int miniPath[][] = new int[triangle.size()][triangle.size()]; 8 for(int i = 0; i < triangle.size(); i++){ 9 for(int j = 0; j < triangle.get(i).size(); j++){ 10 if(0 == i && 0 == j){ 11 miniPath[i][j] = triangle.get(0).get(0); 12 continue; 13 } 14 if(j == 0) 15 miniPath[i][j] = miniPath[i - 1][j] + triangle.get(i).get(j); 16 else if(j == i) 17 miniPath[i][j] = miniPath[i - 1][j - 1] + triangle.get(i).get(j); 18 else 19 miniPath[i][j] = Math.min(miniPath[i - 1][j], miniPath[i - 1][j - 1]) + triangle.get(i).get(j); 20 } 21 } 22 int ret = miniPath[triangle.size() - 1][0]; 23 //showMiniPath(miniPath); 24 for(int i = 1; i < miniPath[0].length; i++){ 25 ret = Math.min(ret, miniPath[triangle.size() - 1][i]); 26 } 27 return ret; 28 } 29 }
这里可以用原来的空间，使得空间复杂度为O（1）
ps:我没有这么做
标签：style blog http io ar color sp for on
原文地址：http://www.cnblogs.com/luckygxf/p/4109548.html
