前言
本系列博客为平时刷LeetCode的记录，每十道题一篇博客，持续更新，所有代码详见GitHub：https://github.com/roguesir/LeetCode-Algorithm
66. Plus One
introduction
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
Example 1:
Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Example 2:
Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Solution
class Solution(object):
def plusOne(self, digits):
"""
:type digits: List[int]
:rtype: List[int]
"""
digits = "".join(map(str, digits))
if digits.isdigit():
return map(int, list(str(int(digits) + 1)))
https://leetcode.com/submissions/detail/202038077/
67. Add Binary
Introduction
Given two binary strings, return their sum (also a binary string).
The input strings are both non-empty and contains only characters 1 or 0.
Example 1:
Input: a = "11", b = "1"
Output: "100"
Example 2:
Input: a = "1010", b = "1011"
Output: "10101"
Solution
class Solution(object):
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
return bin(int(a,2)+int(b,2))[2:]
69. Sqrt(x)
Introduction
Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
Solution
class Solution(object):
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
import math
if x >= 0:
return int(math.sqrt(x))
https://leetcode.com/submissions/detail/202038710/
70. Climbing Stairs
Introduction
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Note: Given n will be a positive integer.
Example 1:
Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. step + 1 step
2. steps
Example 2:
Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. step + 1 step + 1 step
2. step + 2 steps
3. steps + 1 step
Solution
实际上，这道题的结果是斐波那契数列，为了简介代码，我首先想到的是使用lambda表达式，结果，竟然超时了。。。
Method 1
使用lambda表达式构建斐波那契函数。
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
fib = lambda n:1 if n<=2 else fib(n-1)+fib(n-2)
return fib(n+1)
https://leetcode.com/submissions/detail/202039687/
Method 2
直接使用斐波那契数列的通项公式进行计算。
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
from math import sqrt
return int(sqrt(5)/5*((((1+sqrt(5))/2))**(n+1)-(((1-sqrt(5))/2))**(n+1)))
https://leetcode.com/submissions/detail/202040369/
Method 3
递归构建斐波那契函数。
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
a=1
b=1
def fib(n, a, b):
while n != 0:
a, b = b, a+b
n -= 1
return b
return fib(n-1, a, b)
https://leetcode.com/submissions/detail/202222009/
- 更新时间：2019-01-27