update

chapter1/sqrt.py

@@ -2,12 +2,12 @@
 # coding=utf-8
 
 
-################################################################3
+################################################################
 # LeetCode: 69 Sqrt(x)
 # Implement int sqrt(int x).
 # Compute and return the square root of x, where x is guaranteedto be a non-negative integer.
 # Since the return type is an integer, the decimal digits aretruncated and only the integer part of the result is returned.
-################################################################3
+################################################################
 
 class Solution:
     def mySqrt(self, x):

chapter3/1_leetcode7.py

@@ -36,12 +36,12 @@ class Solution:
         3. 将|x|从个位开始向左遍历，依次叠加到结果里面；
         4. 最后判断结果是否溢出（与32位有符号整形的最大值和最小值进行比对）
         """
-        isNegitive = -1 if x < 0 else 1
+        isNegative = -1 if x < 0 else 1
         result, x = 0, abs(x)
         while x > 0:
             result = result * 10 + x % 10
             x = x // 10
-        result = isNegitive * result
+        result = isNegative * result
 
         return 0 if result > INT_MAX_VAL or result < INT_MIN_VAL else result