leetcode--Triangle

1.问题

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.

2.分析

2.1从上往下动态规划，状态转移方程为：

f(i,j) = min{f(i-1,j),f(i-1,j-1)} + (i,j);

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

2.2从右下角往上动态规划，状态转移方程为：

f(i,j)=min{f(i,j+1),f(i+1,j+1)}+(i,j)

// 时间复杂度O(n^2)，空间复杂度O(1)class Solution {public:    int minimumTotal (vector<vector<int>>& triangle) {        for (int i = triangle.size() - 2; i >= 0; --i)            for (int j = 0; j < i + 1; ++j)                triangle[i][j] += min(triangle[i + 1][j],                        triangle[i + 1][j + 1]);        return triangle [0][0];    }};