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042.py
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45 lines (43 loc) · 1.48 KB
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class Solution(object):
def trap(self, height):
"""
:type height: List[int]
:rtype: int
"""
# 按照自己的思路连暴力的都不行;所以最关键一步是从找总体洼地 -> 单个元素计算
# 这是用数组存后半部分最大值的解法
if len(height)<=2:
return 0
n = len(height)
right = [height[-1]] * n
for i in range(n-2, -1,-1):
right[i] = max(height[i], right[i+1])
ans, left = 0, height[0]
for i in range(1, n):
left = max(left, height[i])
ans += min(left, right[i])-height[i]
return ans
# 双指针,比较难想明白。思路是:
# 把上面的数组用 两个元素的直接比较代替了
# i 和 j 位置比较大小;小的那个再和该侧max比看是更新max还是更新
def trap(self, height):
if len(height)<=2:
return 0
n = len(height)
i, j = 0, n-1
left_max, right_max = 0, 0
ans = 0
while i < j:
if height[i] < height[j]:
if height[i] >= left_max:
left_max = height[i]
else:
ans += left_max - height[i]
i += 1
else:
if height[j] >= right_max:
right_max = height[j]
else:
ans += right_max - height[j]
j -= 1
return ans