LeetCode 120. Triangle

[Woodyiiiiiii]

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.

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要注意的是：

1. 三角形数组的创建(C++用vector方便多了)
2. 要按照题目所给的三角形递归路线。

class Solution {
    // stupid problem!!!!!
    // 三角形是按照题目图示所给的三角形
    public int minimumTotal(List<List<Integer>> triangle) {
        int m = triangle.size();
        if (m == 0) {
            return 0;
        }
        if (m == 1) {
            return triangle.get(0).get(0);
        }
        int[][] dp = new int[m][];
        int i = 0, j = 0;
        for (i = 0; i < m; ++i) {
            dp[i] = new int[i + 1];
        }
        dp[0][0] = triangle.get(0).get(0);
        
        for (i = 1; i < m; ++i) {
            for (j = 0; j <= i; ++j) {
                if (j == 0) {
                    dp[i][j] = dp[i - 1][j];
                }
                else if (j == i) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else {
                    dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j - 1]);
                }
                dp[i][j] += triangle.get(i).get(j);
            }
        }
        
        int res = Integer.MAX_VALUE;
        for (j = 0; j < m; ++j) {
            res = Math.min(res, dp[m - 1][j]);
        }
        return res;
    }
}

class Solution {
    // spcae:O(n)
    // Warning:三角形是按照题目图示所给的三角形
    public int minimumTotal(List<List<Integer>> triangle) {
        int m = triangle.size();
        if (m == 0) {
            return 0;
        }
        if (m == 1) {
            return triangle.get(0).get(0);
        }
        int[] dp = new int[m + 1];
        int i = 0, j = 0;
        dp[0] = triangle.get(0).get(0);
        int pre = dp[0];
        
        for (i = 1; i < m; ++i) {
            for (j = 0; j <= i; ++j) {
                if (j == 0) {
                    pre = dp[j];
                }
                else if (j == i) {
                    dp[j] = pre;
                }
                else {
                    int old = dp[j];
                    dp[j] = Math.min(dp[j], pre);
                    pre = old;
                }
                dp[j] += triangle.get(i).get(j);
            }
        }
        
        int res = Integer.MAX_VALUE;
        for (j = 0; j < m; ++j) {
            res = Math.min(res, dp[j]);
        }
        return res;
    }
}

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参考资料:
LeetCode原题