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120. Triangle

120. Triangle

作者: HalcyonMoon | 来源:发表于2016-06-29 11:56 被阅读0次

    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).

    Note:
    Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

    设状态为 f[i][j],表示根到节点(i,j)的最短路径,状态方程:
    f[i][j] = min(f[i-1,j-1],f[i-1,j]) + tringal[i,j]
    使用一维数组实现:

    public class Solution {
        public int minimumTotal(List<List<Integer>> triangle) {
            int height = triangle.size();
            if(height == 0){
                return 0;
            }
            
            int [] minPath = new int[height+1];
            List<Integer> layer = triangle.get(0);
            minPath[0] = Integer.MAX_VALUE;
            minPath[1] = layer.get(0);
            int minPathValue = Integer.MAX_VALUE;
            for(int i = 1; i<height; i++){
                layer = triangle.get(i);
                minPath[i+1] = Integer.MAX_VALUE;
                for(int j=i+1; j>0; j--){
                    minPath[j] = Math.min(minPath[j-1], minPath[j]) + layer.get(j-1);
                }
            }
            for(Integer i: minPath){
                minPathValue = Math.min(minPathValue, i);
            }
            
            return minPathValue;
        }
    }
    

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