【LeetCode】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.

题解：dp问题，题目要求空间不超过o（n）刚开始的想法都用到了n平方的空间，实际上可以把每行n个设置为dp数组，表示改点到根节点的最小和，我的方法由于从上到下需要判断额外边界，实际上从下到上更简单更一般性，而且可以不用额外数组最后的根节点即为最小值

代码：

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

优化：

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

};

最优化：

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