LeetCode 120. Triangle

朴素地递归会超时 —— 有些点的最小路径值被重复计算了。

改用动态规划，dp[i][j]代表自底向上算起，到triangle[i][j]的最小路径值。

代码：

class Solution
{
public:
int minimumTotal(vector<vector<int> > &triangle)
{
auto dp = triangle;

// dp initilization in the first loop
for (int j = 0; j < dp.back().size(); ++ j)
{
dp.back()[j] = triangle.back()[j];
}
for (int i = int(dp.size())-2; i >= 0; -- i)
{
for (int j = 0; j < dp[i].size(); ++ j)
{
dp[i][j] = triangle[i][j] + min(dp[i+1][j], dp[i+1][j+1]);
}
}

return dp[0][0];
}
};

超时的递归解法：
class Solution
{
public:
int minimumTotal(vector<vector<int>> &triangle)
{
return gao(triangle, 0, 0);
}
private:
int gao(vector<vector<int>> &triangle, int x, int y)
{
if (x == triangle.size())
{
return 0;
} else
{
return triangle[x][y] + min(gao(triangle, x+1, y), gao(triangle, x+1, y+1));
}
}
};