LeetCode: Triangle 解题报告

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).

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.

SOLUTION 1:

使用DFS 加记忆矩阵的解法.

mem[i][j]表示第i行第j列的解。它的解可以由下一行推出：mem[i][j] = mem[i+1][j] + mem[i+1][j+1]

/*
    REC, SOL 1:
    */
    public int minimumTotal1(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }
        
        int rows = triangle.size();
        int[][] mem = new int[rows][rows];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < rows; j++) {
                mem[i][j] = Integer.MAX_VALUE;
            }
        }
        
        return dfs(triangle, 0, 0, mem);
    }
    
    public int dfs(List<List<Integer>> triangle, int row, int col, int[][] mem) {
        if (mem[row][col] != Integer.MAX_VALUE) {
            return mem[row][col];
        }
        
        if (row == triangle.size() - 1) {
            mem[row][col] = triangle.get(row).get(col);
        } else {
            int left = dfs(triangle, row + 1, col, mem);
            int right = dfs(triangle, row + 1, col + 1, mem);    
            mem[row][col] = triangle.get(row).get(col) + Math.min(left, right);
        }
        
        return mem[row][col];
    }

SOLUTION 2:

ref: http://blog.csdn.net/imabluefish/article/details/38656211

动态规划的题目

我们可以轻松将上面的修改为DP.

并且，为了减少内存使用量，使用一维DP即可。

f[j] 表示下一行第j列某点到最后底部的最短值。因为我们只需要下一行的这个值，所以我们使用一行的DP memory即可完成任务。

第一步： 先计算出最后一排的最短值，实际上就是这一排本身的值。
第二步：From bottom to up, 每一层的最短值只需要把自身值加上，并且取下层的左右邻接点的最小值。

/*
    DP, SOL 2:
    */
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }
        
        int rows = triangle.size();
        int[] D = new int[rows];
        
        for (int i = rows - 1; i >= 0; i--) {
            // 注意：边界条件是 j <= i
            for (int j = 0; j <= i; j++) {
                if (i == rows - 1) {
                    D[j] = triangle.get(i).get(j);
                } else {
                    D[j] = triangle.get(i).get(j) + Math.min(D[j], D[j + 1]);
                }    
            }
        }
        
        return D[0];
    }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dp/MinimumTotal.java