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社区首页 >专栏 >【leetcode】Triangle

【leetcode】Triangle

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阳光岛主
发布2019-02-19 11:34:53
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发布2019-02-19 11:34:53
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文章被收录于专栏:米扑专栏米扑专栏

Question:

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

代码语言:javascript
复制
[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

Anwser 1:

代码语言:javascript
复制
class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int rows = triangle.size();
        if(0 == rows) return 0;
        
        int *minSums = new int[rows];
        int *temp = new int[rows];
        
        for(int r = 0; r < rows; r++) {
            vector<int> vec = triangle[r];
            temp[0] = vec[0] + (r > 0 ? minSums[0] : 0);
            for(int i = 1; i < r; i++) {
                temp[i] = vec[i] + min(minSums[i-1], minSums[i]);
            }
            
            if(r > 0) {
                temp[r] = vec[r] + minSums[r-1];
            }
                
            int *tswap = temp;
            temp = minSums;
            minSums = tswap;
        }
        
        int m = minSums[0];
        for(int i = 1; i < rows; i++) 
        {
            if(minSums[i] < m){
                m = minSums[i];
            } 
        }
        
        delete temp;
        delete minSums;
        return m;
    }
};

Anwser 2:  

代码语言:javascript
复制
class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int line = triangle.size();

        for(int i = line -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]);
            }
        }

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

Anwser 3:

代码语言:javascript
复制
class Solution {
public:
    
    void run(vector<vector<int> > &triangle, int row, int idx, int curSum, int &minPath)
    {
        if (row == triangle.size())
        {
            minPath = min(minPath, curSum);
            return;
        }
        
        run(triangle, row + 1, idx, curSum + triangle[row][idx], minPath);
        run(triangle, row + 1, idx + 1, curSum + triangle[row][idx], minPath);
    }

    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int minPath = INT_MAX;
        
        if (triangle.size() == 0) {
            return 0;
        }
        
        run(triangle, 0, 0, 0, minPath);
        
        return minPath;
    }
};

注意点:

1) Judge Small is ok, but Judge Large is error

2) 如果迭代很深的话,容易造成压栈占用内存很高,超出段后会溢出

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原始发表:2013年04月14日,如有侵权请联系 cloudcommunity@tencent.com 删除

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