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2014
11-18

LeetCode-Triangle[动态规划]

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.

标签: Array Dynamic Programming
分析

设状态为$f(i, j)$,表示从从位置$(i,j)$出发,路径的最小和,则状态转移方程为
$$
f(i,j)=\min\left\{f(i,j+1),f(i+1,j+1)\right\}+(i,j)
$$

代码1

// LeetCode, Triangle
// 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];
    }
};

Java代码:

class Solution:
    # @param triangle, a list of lists of integers
    # @return an integer
    def minimumTotal(self, triangle):
        n = len(triangle)
        arr = [0]*n
        tmpArr = [0]*n
        arr[0] = triangle[0][0]
        for i in range(1,n):
            tmpArr[0] = arr[0] + triangle[i][0]
            for j in range(1,i):
                tmpArr[j] = min(arr[j-1],arr[j]) + triangle[i][j]
            tmpArr[i] = arr[i-1] + triangle[i][i]
            tmp = tmpArr
            tmpArr = arr
            arr = tmp
        return min(arr)
        

  1. 您没有考虑 树的根节点是负数的情况, 若树的根节点是个很大的负数,那么就要考虑过不过另外一边子树了