本文共 3263 字，大约阅读时间需要 10 分钟。

• ：字符串转置，Easy

  定义头尾指针，调换对应的字符

class Solution(object):    def reverseString(self, s):        """        :type s: str        :rtype: str        """        s = list(s)        start = 0        end = len(s)-1        while(start

• ：删除排序数组中的重复数字，返回剩余数组的长度，Easy

class Solution(object):    def removeDuplicates(self, nums):        if nums == []: return 0        j = 1        for i in range(1,len(nums)):            if nums[i]!=nums[i-1]:                nums[j] = nums[i]                j+=1        return j

• ：删除数组中指定的数字，返回剩余数组的长度，Easy

class Solution(object):    def removeElement(self, nums, val):        """        :type nums: List[int]        :type val: int        :rtype: int        """        if nums == []: return 0        j = 0        for i in range(len(nums)):            if nums[i]!=val:                nums[j]=nums[i]                j+=1        return j

• ：将数组中的0移到最后，Easy

  用j表示当前可填充的位置

class Solution(object):    def moveZeroes(self, nums):        j=0        for i in range(len(nums)):            if nums[i]!=0:                nums[j]=nums[i]                j=j+1        for i in range(j,len(nums)):            nums[i]=0

• ：输入数组有序，返回和等于target的两个数的下标

  二分查找

class Solution(object):    def twoSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[int]        """        for i in range(len(nums)):            tmp=target-nums[i]            start=i+1            end=len(nums)-1            while(start<=end):                if tmp==nums[end]:                    return [i+1,end+1]                if tmp==nums[start]:                    return [i+1,start+1]                mid = (end - start) / 2+start                if mid==start or mid==end:                    break                if tmp > nums[mid]:                    start=mid                elif tmp< nums[mid]:                    end=mid                else:                    return [i+1,mid+1]

• ：逆转字符串中的元音字母，Easy

class Solution(object):    def reverseVowels(self,s):        dic=['a','o','e','i','u','A','O','E','I','U']        tmp=list(s)        start,end = 0,len(tmp)-1        while(start
   
    start and tmp[end] not in dic):                end-=1            if start
   

• ：分割字符串，使得每个字母只会在一个子字符串中出现一次，Medium

Greedy

用字典存储字符第一次出现的位置，遍历字符串，如果当前字符已经出现过，则意味着要合并到第一次出现的字符所在的区间中

class Solution(object):    def partitionLabels(self, S):        """        :type S: str        :rtype: List[int]        """        stack = []        maps = dict()        for i,s in enumerate(S):            if s not in maps:                maps[s] = i                stack.append([i,i])            else:                idx = maps[s]                while(stack):                    [start,end] = stack.pop()                    if idx>=start:                        break                stack.append([start,i])        ret = []        for i in stack:            ret.append(i[1]-i[0]+1)        return ret

• ：给定字符串s和列表d,判断s删去一部分字符是否可以组成d中的字符串，如果可以求长度最长且字典序最小的字符串。否则返回空串。Medium

class Solution(object):    def findLongestWord(self, s, d):        """        :type s: str        :type d: List[str]        :rtype: str        """        maxlen = [len(token) for token in d]        pos = [0 for token in d]        ret = []        for token in s:            for i in range(len(pos)):                if pos[i]

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