leetcode 120. 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.

思路：动态规划经典题。

解法一：动态规划顺推解法

 int minimumTotal(vector<vector<int>>& T) {
        int r = T.size();
        if (r == 0)  return 0;
        vector<vector<int>> dp(r, vector<int>(r, 0));
        dp[0][0] = T[0][0];
        for (int i = 1; i < r; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (j == 0) dp[i][j] = dp[i - 1][j] + T[i][j];
                else if (j == i) dp[i][j] = dp[i - 1][j - 1] + T[i][j];
                else dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j]) + T[i][j];
            }
        }
        int minimum = dp[r - 1][0];
        for (int j = 1; j < r; ++j)
            minimum = min(minimum, dp[r - 1][j]);
        return minimum;
    }

时间复杂度：$O(n^2)$，空间复杂度：$O(n^2)$

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解法二：动态规划逆推解法

    int minimumTotal(vector<vector<int>>& T) {
        int len = (T.size() > 0 ? T.size() : 0);
        if (len == 0) return 0;
        vector<int> dp(len, 0);
        for (int j = 0; j < len; ++j) dp[j] = T[len - 1][j];
        for (int i = len - 2; i >= 0; --i) {
            for (int j = 0; j <= i; ++j) {
                dp[j] = min(dp[j], dp[j + 1]) + T[i][j];
            }
        }
        return dp[0];
    }

时间复杂度：$O(n^2)$，空间复杂度：$O(n)$