【medium】120. 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.

 

这个写的很好： https://www.cnblogs.com/mozi-song/p/9615167.html

从两个角度来解题：

（1）top bottom (memorization)

 

class Solution {
private:
    int minimum(vector<vector<int>>& triangle, vector<vector<int>>& dp, int i, int j){
　　　　　// 1.边界
        if (i == triangle.size()-1){
            dp[i][j] = triangle[i][j];
            return triangle[i][j];
        }
        // 2.如果cached
        if (dp[i][j] != INT_MAX)
            return dp[i][j];
        // 3.递推公式
        dp[i][j] = triangle[i][j] + min({minimum(triangle, dp, i+1, j), minimum(triangle, dp, i+1, j+1)});
        
        return dp[i][j];
    }
public:
    int minimumTotal(vector<vector<int>>& triangle) {

        vector<vector<int>> dp;  // 1.记忆数组
        
　　　　　//2.记忆数组初始化
        dp.resize(triangle.size());//r行
        for (int k = 0; k < triangle.size(); k++){
           dp[k].resize(triangle.size());//每行为c列
        }
        for (int i=0; i<triangle.size(); i++){
            for (int j=0; j<triangle.size(); j++)
                dp[i][j] = INT_MAX;
        }
   　　　//3. solution函数 
        minimum(triangle, dp, 0, 0);
        return dp[0][0];
    }
};

  (2) bottom up (tabulation) --- 更加简单直观

 

class Solution{
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if (triangle.size() == 1)
            return triangle[0][0];
        
        vector<vector<int>> dp;
        dp.resize(triangle.size());//r行
        for (int k = 0; k < triangle.size(); k++){
           dp[k].resize(triangle.size());//每行为c列
        }
        
        for (int i=triangle.size()-1; i>=0; i--)
            dp[triangle.size()-1][i] = triangle[triangle.size()-1][i];
        
        for (int i=triangle.size()-2; i>=0; i--){
            for (int j=0; j<=i; j++)
                dp[i][j] = triangle[i][j] + min({dp[i+1][j], dp[i+1][j+1]});
        }
        
        return dp[0][0];
    }
};