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.

$$f(i,j)=\min\left\{f(i,j+1),f(i+1,j+1)\right\}+(i,j)$$

// 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