[LeetCode] 120, and minimum path triangle

Title Description

Given a triangle, find the minimum and the top-down path. Each step can move to the next line adjacent nodes.

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]
自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。请使用O(n)的空间。

Problem-solving ideas

This question is not difficult, not to be imprisoned idea, in fact, you can use $O (1)$ extra space problem solving.

• Top-down: We can use the original vector<vector<int>>& triangleitself is stored state, the case space complexity $O (1)$ .
• Bottom-up : sometimes change the direction of the problem may be more simple, then you can not consider a variety of boundary conditions. (I realize)

Reference Code

I realize (top-down)

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int rows = triangle.size();
        for(int i = 1; i < rows; i++){
            for(int j = 0; j <= i; j++){
                int left = j-1 >= 0? triangle[i-1][j-1]: INT_MAX;
                int right = j <= i-1? triangle[i-1][j]: INT_MAX;
                triangle[i][j] = min(left, right) + triangle[i][j];
            }
        }
        
        int res = INT_MAX;
        for(int i = 0; i < rows; i++)
            res = min(res, triangle[rows-1][i]);
        return res;
    }
};

Big brother implement (bottom-up)

class Solution {
public:
    //题解1：动态规划，自底向上，从倒数第二层开始向上求解（不用边界处理）
    //状态转移方程为：triangle[i][j]+=min(triangle[i+1][j],triangle[i+1][j+1])
    //时间复杂度为O(m*n)，空间复杂度为O(1)
    int minimumTotal_1(vector<vector<int>>& triangle) {
        for(int i=triangle.size()-2;i>=0;--i){
            for(int j=0;j<triangle[i].size();++j){//每个点的路径和等于该点的值加上它正下面和右下中较小点的值
                triangle[i][j]+=min(triangle[i+1][j],triangle[i+1][j+1]);
            }
        }
        return triangle[0][0];
    }

    //题解2：降维变为一维dp，自底向上，因为其实每次只会用到上一层数据，因此不需二维数组存储所有可能情况来进行比较
    //也就是说一维dp[j]是随着i进行覆盖更新上一层的dp[j]
    int minimumTotal(vector<vector<int>>& triangle){
        int n=triangle.size(),dp[n+1];
        memset(dp,0,sizeof(dp));
        for(int i=n-1;i>=0;--i){
            for(int j=0;j<=i;++j){//第i层只有i列
                dp[j]=min(dp[j],dp[j+1])+triangle[i][j];
            }
        }
        return dp[0];
    }
};