2021-08-14 数组：搜索遍历
乘积最大子数组：动归
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挖掘几点信息：
- 除自身以外：即左边乘积*右边乘积
- 乘积：累乘，可以利用先前结果（用两个变量维护）
- 记录：直接使用答案数组记录，左至右左边乘积，反之亦然，然后乘以变量即为累积的另一边的成绩。故而至少需要两个数组，无法进行原地覆盖。
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [1]*n
per, suf = 1, 1
for i in range(n):
ans[i] = ans[i] * per
ans[n-i-1] = ans[n-i-1] * suf
per = per * nums[i]
suf = suf * nums[n-i-1]
return ans
递增的三元子序列
- 贪心算法：得到一个就霸占一个，直到符合条件为止
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
one = two = float('inf')
for three in nums:
if three > two: return True
elif three <= one: one = three
else: two = three
return False;
搜索二维矩阵Ⅱ
- 二分法（对应二维的就是划分为四个矩阵，左侧两个矩阵中可能存在比当前元素小的，故而顺着寻找列即可）：
class Solution:
def searchMatrix(self, matrix, target):
# an empty matrix obviously does not contain `target`
if not matrix:
return False
def search_rec(left, up, right, down):
if left > right or up > down:
return False
elif target < matrix[up][left] or target > matrix[down][right]:
return False
mid = left + (right-left)//2
# Locate `row` such that matrix[row-1][mid] < target < matrix[row][mid]
row = up
while row <= down and matrix[row][mid] <= target:
if matrix[row][mid] == target:
return True
row += 1
return search_rec(left, row, mid-1, down) or search_rec(mid+1, up, right, row-1)
return search_rec(0, 0, len(matrix[0])-1, len(matrix)-1)
- 减枝：通过循迹的方式完成
class Solution:
def searchMatrix(self, matrix, target):
# an empty matrix obviously does not contain `target` (make this check
# because we want to cache `width` for efficiency's sake)
if len(matrix) == 0 or len(matrix[0]) == 0:
return False
# cache these, as they won't change.
height = len(matrix)
width = len(matrix[0])
# start our "pointer" in the bottom-left
row = height-1
col = 0
while col < width and row >= 0:
if matrix[row][col] > target:
row -= 1
elif matrix[row][col] < target:
col += 1
else: # found it
return True
return False
