方法一:利用一个字典和一个列表解决,字典记录每个字符出现的次数,列表记录出现过的字符,从前到后遍历这个string 遇到新字符(字典中没有的key),在字典中添加key为该字符,value为1,列表中append该字符 遇到出现过的字符(字典中有该key),字典中该key的value加1 然后,遍历列表,打印出value为1的字符即可 代码如下 1 ...
public static void main(String[] args) { // Declare and initialize a string variable. String str1 = "gibblegabbler"; // Print the original string. System.out.println("The given string is: " + str1); // Loop through each character of the string. for (int i = 0; i < str1.lengt...
Given a string "teeter", the first non repeating character would be 'r'. in "toothless", it would be 'h'. I'm wondering about the most efficient way to get this done? One option is to use a hash table, with the characters in the string as keys, and frequencies of each character...
You are given a string str consisting of lowercase Latin letters. Find the first non-repeating character in str. Note: You have to traverse the string only once. See original problem statement here Tets Case: Input: prepbytes Output: 1 Explanation: In the string 'prepbytes', we start trave...
First non-repeating character in a stream Given an input stream of n characters consisting only of small case alphabets the task is to find the first non repeating character each time a character is inserted to the stream. Example Flow in stream : a, a, b, c ...
The method may include identifying locations in a mask value to store the unique random value and locations in a mask value to store filler values. The mask value may include different character sets and may comply with user specified formats. The method may include generating a set of ...
[Algorithm] 387. First Unique Character in a String 2019-12-13 22:36 −Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0... ...
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[Algorithm] 387. First Unique Character in a String 2019-12-13 22:36 −Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0... ...
iflen(c)>=2{b:=make([]int,len(c))copy(b,c)*res=append(*res,b)}visited:=map[int]bool{}fori:=current+1;i<len(nums);i++{ifnums[current]<=nums[i]{if_,ok:=visited[nums[i]];ok{continue}else{visited[nums[i]]=truegenerateIncSubsets(nums,i,c,res)}}}c=c[:len(c)-1]...