[leetcode.com]算法题目

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.

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
    }
};

思路：假设三角形共有n行，题目中看出第i行共有i个元素。从top到bottom，我们只考虑minimum path的最后一步是停在了这n个元素的哪一个上面。用数组min_sum(k)表示最后一行到达第k个元素的最小路径（min_sum总长度为n），然后找出min_sum(k)中最小的元素即可。注意一下自上而下递推min_sum时候的递推公式，代码如下：

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        
        int n = triangle.size();
        if (0 == n) return 0;
        if (1 == n) return triangle[0][0];
        
        int *min_sum = new int[n];
        for(int i=0; i<n;i++)
            min_sum[i] = 0;
        
        min_sum[0]=triangle[0][0];
        for(int i=1;i<n;i++){
            for(int j=triangle[i].size()-1; j>=0;j--){
                if (0==j){
                    min_sum[j] += triangle[i][0]; 
                }else if (triangle[i].size()-1==j){
                    min_sum[j] = min_sum[j-1]+triangle[i][j];
                }else{
                    min_sum[j] = min(min_sum[j-1], min_sum[j])+triangle[i][j];
                }
            }
        }
        
        int minTotal = min_sum[0];
        for(int i=1;i<n;i++){
            minTotal = min(minTotal, min_sum[i]);
        }
        
        delete[] min_sum;
        return minTotal;
    }
    
    int min(int a, int b){
        return (a>b?b:a);
    }
};

注意：一开始的时候内部j那个循环是正着写，后来发现这样有个问题，就是有可能会使用已经变动的过的值去更新下一列。