Triangle

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Triangle

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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.

思路1：搜索题，找出从顶到底路径中最小的和，则要在所有路径中即可行解中找出最优解。

直接用dfs查找TLE，因为有很多路径重复计算了。

class Solution {public:    int minimumTotal(vector<vector<int> > &triangle) {        if (triangle.empty()) return 0;        ans = INT_MAX;        dfs(triangle, 1, 0, triangle[0][0]);        return ans;    }private:    int ans;    void dfs(vector<vector<int> > &triangle, int depth, int col, int sum) {        // update solution        if (depth == triangle.size()) {            if (sum < ans)                ans = sum;            return;        }        dfs(triangle, depth + 1, col, sum + triangle[depth][col]);        dfs(triangle, depth + 1, col + 1, sum + triangle[depth][col + 1]);    }};

思路2：为了避免重复计算，采用备忘录，用一个数组保存那个位置到底端的最优解。

class Solution {public:    int minimumTotal(vector<vector<int> > &triangle) {        if (triangle.empty()) return 0;        for (int i = 0; i < triangle.size(); ++i) {            memory.push_back(vector<int>(triangle[i].size(), INT_MAX));        }        return find(triangle, 0, 0);    }private:    vector<vector<int> > memory;    int find(vector<vector<int> > &triangle, int depth, int col) {        if (memory[depth][col] != INT_MAX) {            return memory[depth][col];        }                if (depth == triangle.size() - 1) {            return memory[depth][col] = triangle[depth][col];        }        int left_sum = find(triangle, depth + 1, col);        int right_sum = find(triangle, depth + 1, col + 1);                int min_sum = min(left_sum, right_sum);        min_sum += triangle[depth][col];                return memory[depth][col] = min_sum;    }};

思路3：动态规划，从底向上，存储到低端的最优解，求解当前最优解时，只需用到下一层相邻节点的最优解。

方程：dp[i][j] = min(dp[i+1][j], dp[i+1][j+1]) + triangle[i][j] ( 0 <= i < max_depth-1)。

class Solution {public:    int minimumTotal(vector<vector<int> > &triangle) {        if (triangle.empty()) return 0;        for (int i = 0; i < triangle.size(); ++i) {            dp.push_back(vector<int>(triangle[i].size(), 0));        }        int depth = triangle.size() - 1;        int col = 0;        for ( ; col < triangle[depth].size(); ++col) {            dp[depth][col] = triangle[depth][col];        }        for (depth = triangle.size() - 2; depth >= 0; --depth) {            for (col = 0; col < triangle[depth].size(); ++col) {                dp[depth][col] = min(dp[depth + 1][col], dp[depth + 1][col + 1]) + triangle[depth][col];            }        }        return dp[0][0];    }private:    vector<vector<int> > dp;};

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