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‎leetcode/012.整数转罗马数字.py

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class Solution(object):
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def intToRoman(self, num):
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res = ''
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d = {'M': 1000, 'CM': 900, 'D': 500, 'CD': 400, 'C': 100, 'XC': 90,
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'L': 50, 'XL': 40, 'X': 10, 'IX': 9, 'V': 5, 'IV': 4, 'I': 1}
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# 字典无序，需要先排序，再逆转
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for char, val in sorted(d.items(), key = lambda x: x[1])[::-1]:
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while num >= val:
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res += char
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num -= val
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return res

‎leetcode/013.罗马数字转整数.py

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# 字符的值比后面小则可以合并，即先减当前数，再加后面数
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class Solution(object):
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def romanToInt(self, s):
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d = {'I':1, 'V':5, 'X':10, 'L':50, 'C':100, 'D':500, 'M':1000}
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res = 0
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for i in range(len(s)):
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if i<len(s)-1 and d[s[i]] < d[s[i+1]]:
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res -= d[s[i]]
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else:
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res += d[s[i]]
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return res

‎leetcode/014.最长公共前缀.py

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# 方法一
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class Solution(object):
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def longestCommonPrefix(self, strs):
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if not strs:
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return ''
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# 遍历索引
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for i in range(len(strs[0])):
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# 遍历每个字符串
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for str in strs:
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# 索引超过字符串长度或字符不相等
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if len(str) <= i or strs[0][i] != str[i]:
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return strs[0][:i]
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return strs[0]
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# 方法二：使用python函数
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class Solution(object):
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def longestCommonPrefix(self, strs):
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return os.path.commonprefix(strs)

‎leetcode/017.电话号码的字母组合.py

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# 递归
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class Solution(object):
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def letterCombinations(self, digits):
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d = {'2':'abc', '3':'def', '4':'ghi', '5':'jkl', '6':'mno', '7':'pqrs', '8':'tuv', '9':'wxyz'}
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def combine(s, digits):
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# 终止条件，数字为空时把字符串添加到数组中
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if not digits:
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res.append(s)
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else:
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for char in d[digits[0]]:
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combine(s+char, digits[1:])
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res = []
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if not digits:
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return res
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combine('', digits)
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return res

‎leetcode/020.有效的括号.py

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# 利用栈找对应括号，遇到左括号则进栈，遇到右括号则出栈，判断两个括号是否匹配
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class Solution(object):
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def isValid(self, s):
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left, right = '([{', '}])'
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stack = []
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for i in s:
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if i in left:
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stack.append(i)
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if i in right:
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if not stack:
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return False
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res = stack.pop()
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if (i==')' and res!='(') or (i==']' and res!='[') or (i=='}' and res!='{'):
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return False
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return not stack

‎leetcode/022.括号生成.py

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class Solution(object):
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def generateParenthesis(self, n):
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self.res = []
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self.generate('', 0, 0, n)
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return self.res
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def generate(self, s, left, right, n):
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# 左右括号用完时，添加该括号字符串
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if left == n and right == n:
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self.res.append(s)
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# 左括号未完时，添加左括号，再递归
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if left < n:
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self.generate(s+'(', left+1, right, n)
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# 右括号少于左括号时，可添加右括号，再递归
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if right < left:
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self.generate(s+')', left, right+1, n)

‎leetcode/026.删除排序数组中的重复项.py

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# 遍历数组，遇到不重复的值就按顺序复制到前面
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class Solution(object):
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def removeDuplicates(self, nums):
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if not nums:
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return 0
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j = 0
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for i in range(len(nums)):
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if nums[i] != nums[j]:
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nums[j+1] = nums[i]
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j += 1
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return j+1

‎leetcode/027.移除元素.py

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# 将非val的值复制到数组前，最后一个复制完后，返回的索引即前面数组的长度
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class Solution(object):
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def removeElement(self, nums, val):
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j = 0
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for i in range(len(nums)):
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if nums[i] != val:
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nums[j] = nums[i]
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j += 1
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return j

‎leetcode/028.实现strStr().py

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# 方法一
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class Solution(object):
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def strStr(self, haystack, needle):
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if not needle:
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return 0
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elif needle not in haystack:
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return -1
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else:
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return haystack.index(needle)
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# 方法二
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class Solution(object):
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def strStr(self, haystack, needle):
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return haystack.find(needle)

‎leetcode/035.搜索插入位置.py

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class Solution(object):
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def searchInsert(self, nums, target):
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if target in nums:
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return nums.index(target)
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for i in range(len(nums)):
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if nums[i] > target:
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return i
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return i+1

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