LeetCode 120. Triangle

动态规划 问题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.

自顶向下 递归+记忆化

时间复杂度O(n^2) 空间复杂度O(n^2)

class Solution {
public:
    
    int helper(vector<vector<int>>& triangle,int x,int y,vector<vector<int>>& dp) {
        int n = triangle.size();
        if(x>=n||y>=n) return 0;
        
        if(dp[x][y])
            return dp[x][y];
        else
            dp[x][y] = min(helper(triangle,x+1,y,dp),helper(triangle,x+1,y+1,dp)) + triangle[x][y];
        return dp[x][y];
        
    }
    
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        if(!n) return 0;
        
        vector<vector<int>> dp(n,vector<int>(n,0));
        int res = helper(triangle,0,0,dp);
        return res;
    }
};

自底向上优化空间到o(n)

class Solution {
public:
    
    int helper(vector<vector<int>>& triangle,int x,int y,vector<int>& dp) {
        int n = triangle.size();
        if(x>=n||y>=n) return 0;
        for(int i=0;i<n;++i)
            dp[i] = triangle[n-1][i];
        
        for(int i=n-2;i>=0;--i) {
            for(int j=0;j<=i;++j) {
                dp[j] = min(dp[j],dp[j+1]) + triangle[i][j];
            }
        }
        
        return dp[0];
        
    }
    
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        if(!n) return 0;
        
        vector<int> dp(n,0);
        int res = helper(triangle,0,0,dp);
        return res;
    }
};