LeetCode 120. Triangle [python3/C++]

一 题目

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.

二 分析

    题目大意是给定一个三角形。从三角形顶点找一条路径到底部，使得该路径上的值加起来的值最小。

    明显可以看出，最左边的点只能从右上方经过，最右边的点只能从左上方经过，而中间的点可能从左上方或者右上方的点经过。于是从第一层到最后一层进行dp，更新到达这个点的最小路径权值，最后遍历一遍底层的值，最小的就是答案。

三 代码

python3：120. Triangle

class Solution:
    def minimumTotal(self, triangle):
        """
        :type triangle: List[List[int]]
        :rtype: int
        """
        n = len(triangle)
        dp =  [[0 for col in range(n)] for row in range(n)]
        dp[0][0] = triangle[0][0]
        ans = 200000000
        for i in range(1, n):
            for j in range(i + 1):
                if j == 0:
                    dp[i][j] = dp[i-1][0] + triangle[i][j]
                elif j == i:
                    dp[i][j] = dp[i-1][j-1] + triangle[i][j]
                else:
                    dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]) + triangle[i][j]
        for i in range(n):
            ans = min(ans, dp[n-1][i])
        return ans

C++：

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        int dp[n+2][n+2], ans = 2000000000;
        dp[0][0] = triangle[0][0];
        for(int i = 1; i < n; i++) {
            for(int j = 0; j <= i; j++) {
                if(j == 0)
                    dp[i][j] = dp[i-1][0] + triangle[i][j];
                else if(j == i)
                    dp[i][j] = dp[i-1][j-1] + triangle[i][j];
                else
                    dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]) + triangle[i][j];
            }
        }
        for(int i = 0; i < n; i++) {
            ans = min(ans, dp[n-1][i]);
        }
        return ans;
    }
};