[leetcode.com]算法题目 - 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.

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

     }
 };

答题模板

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 ( == n) return ;
         if ( == n) return triangle[][];

         int *min_sum = new int[n];
         for(int i=; i<n;i++)
             min_sum[i] = ;

         min_sum[]=triangle[][];
         for(int i=;i<n;i++){
             for(int j=triangle[i].size()-; j>=;j--){
                 if (==j){
                     min_sum[j] += triangle[i][];
                 }else if (triangle[i].size()-==j){
                     min_sum[j] = min_sum[j-]+triangle[i][j];
                 }else{
                     min_sum[j] = min(min_sum[j-], min_sum[j])+triangle[i][j];
                 }
             }
         }

         int minTotal = min_sum[];
         for(int i=;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);
     }
 };

Answer

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。