此题与jz067一样。这个算法的时间复杂度是O(N),空间复杂度是O(1)。
class Solution:
def strToInt(self, str: str) -> int:
res, i, sign, length = 0, 0, 1, len(str)
int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10
if not str: return 0 # 空字符串,提前返回
while str[i] == ' ':
i += 1
if i == length: return 0 # 字符串全为空格,提前返回
if str[i] == '-': sign = -1
if str[i] in '+-': i += 1
for c in str[i:]:
if not '0' <= c <= '9' : break
if res > bndry or res == bndry and c > '7':
return int_max if sign == 1 else int_min
res = 10 * res + ord(c) - ord('0')
return sign * res
PREVIOUSjz068I.二叉搜索树的最低公共祖先