Triangle

Triangle

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

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if(null == triangle)
            return 0;
        if(1 == triangle.size())
            return triangle.get(0).get(0);
        int miniPath[][] = new int[triangle.size()][triangle.size()];
        for(int i = 0; i < triangle.size(); i++){
            for(int j = 0; j < triangle.get(i).size(); j++){
                if(0 == i && 0 == j){
                    miniPath[i][j] = triangle.get(0).get(0);
                    continue;
                }                    
                if(j == 0)
                    miniPath[i][j] = miniPath[i - 1][j] + triangle.get(i).get(j);
                else if(j == i)
                    miniPath[i][j] = miniPath[i - 1][j - 1] + triangle.get(i).get(j);
                else
                    miniPath[i][j] = Math.min(miniPath[i - 1][j], miniPath[i - 1][j - 1]) + triangle.get(i).get(j);
            }
        }
        int ret = miniPath[triangle.size() - 1][0];
        //showMiniPath(miniPath);
        for(int i = 1; i < miniPath[0].length; i++){
            ret = Math.min(ret, miniPath[triangle.size() - 1][i]);
        }
        return ret;
    }
}

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