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.
題解:
依然是動態規劃題。
記錄到每一層的每一個位置所需要的最小步數,層層遍歷下去即可。其具體公式如下。
對於每一層,不爲0和最後1個的那一步i和當前層level,其計算方式爲。
step[level][i] = min( step[level-1][i-1] + triangle[level][i], step[level-1][i] + triangle[level][i])
當爲第一個時,那麼只有從上一層的0到達,最後一個時,同理。
當然題目說如果能用O(n)的空間是最好的。因此又使用了一個滾動數組。
具體做法是開2個數組。分別表示到達這一層和上一層的最小步數,代碼如圖。
step = [[999999 for i in range(len(triangle))], [999999 for i in range(len(triangle))]]
而狀態判定也需要修改一下,改爲使用index判斷而不是level。index用level % 2來維護。index-1所對應的便是到達上一層的最小步數。
對於不是最後一個和不是第一個的元素i,修改如下
step[index][i] = min(step[index - 1][i - 1] + triangle[level][i], step[index - 1][i] + triangle[level][i])
代碼如下
class Solution(object):
def minimumTotal(self, triangle):
if len(triangle) == 0:
return 0
if len(triangle) == 1:
return triangle[0][0]
step = [[999999 for i in range(len(triangle))], [999999 for i in range(len(triangle))]]
level = 1
step[0][0] = triangle[0][0]
while level < len(triangle):
index = level % 2
i = 0
while i < len(triangle[level]):
if i > 0 and i < len(triangle[level]) - 1:
step[index][i] = min(step[index - 1][i - 1] + triangle[level][i],
step[index - 1][i] + triangle[level][i])
# tmplist[i] = min(step[i] + triangle[level][i] , step[i-1] + triangle[level][i])
else:
if i == 0:
step[index][i] = step[index - 1][i] + triangle[level][i]
else:
step[index][i] = step[index - 1][i - 1] + triangle[level][i]
i = i + 1
level = level + 1
return min(step[index])