LeetCode_Triangle

最新推荐文章于 2025-08-09 16:37:48 发布

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最新推荐文章于 2025-08-09 16:37:48 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/graph/p/3251518.html

本文介绍如何使用动态规划解决最小路径和问题。通过分析三角形中从顶点到底边的最短路径总和，展示了算法的实现过程，并提供了一个C/C++代码示例。重点在于理解动态规划在解决此类问题时的优势，以及如何优化空间复杂度。

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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.

　　分析：This problem is more likely to be a (dynamic programming) DP problem,
where a[n][i] = a[n][i]+min(a[n-1][i], a[n-1][i-1]).
Note that in this problem, "adjacent" of a[i][j] means a[i-1][j] and a[i-1][j-1], if available(not out of bound), while a[i-1][j+1] is not "adjacent" element.

The minimum of the last line after computing is the final result.

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = triangle.size();
        for( int i = 1; i< len ; i++)
          for( int j = 0; j < triangle[i].size(); j++){
            if(j == 0){
                triangle[i][j] += triangle[i-1][j];
                continue;
            }
            if(j == triangle[i].size() -1){
                triangle[i][j] += triangle[i-1][j-1];
                continue;
            }
            int val = triangle[i-1][j] < triangle[i-1][j-1] ? triangle[i-1][j] 
                                            :triangle[i-1][j-1] ;
           triangle[i][j] += val;
          }
          
        int minval = triangle[len-1][0];
        
        for(int i = 1 ; i< triangle[len-1].size() ; i++){
            if(minval > triangle[len-1][i] ) minval = triangle[len-1][i];
        }
        
        return minval;
    }
};