leetcode-Triangle

Difficulty: Medium

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, wheren is the total number of rows in the triangle.

第一种方法，递归，超时

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

第二种方法，通过，但空间复杂度不理想

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

第三种优化

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