Triangle

描述

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.

分析

设状态为f(i, j)，表示从从位置(i,j)出发，路径的最小和，则状态转移方程为

$$f(i,j)=\min\left\{f(i+1,j),f(i+1,j+1)\right\}+(i,j)$$

代码

Python

// Triangle
// 时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        for (int i = triangle.size() - 2; i >= 0; --i)
            for (int j = 0; j < i + 1; ++j) {
                int old = triangle.get(i).get(j);
                triangle.get(i).set(j, old + Math.min(triangle.get(i + 1).get(j),
                        triangle.get(i + 1).get(j + 1)));
            }

        return triangle.get(0).get(0);
    }
}

Java

// Triangle
// 时间复杂度O(n^2)，空间复杂度O(1)
class Solution {
public:
    int minimumTotal (vector<vector<int>>& triangle) {
        for (int i = triangle.size() - 2; i >= 0; --i)
            for (int j = 0; j < i + 1; ++j)
                triangle[i][j] += min(triangle[i + 1][j],
                        triangle[i + 1][j + 1]);

        return triangle [0][0];
    }
};

C++

# Triangle
# 时间复杂度O(n^2)，空间复杂度O(1)
class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        for i in range(len(triangle) - 2, -1, -1):
            for j in range(i + 1):
                old = triangle[i][j]
                triangle[i][j] = old + min(triangle[i + 1][j],
                                         triangle[i + 1][j + 1])
        return triangle[0][0]