[抄题]:
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.
Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.
Follow up:
Could you solve it using only O(1) extra space?
Example 1:
Input: ["a","a","b","b","c","c","c"] Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"] Explanation: "aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
Example 2:
Input: ["a"] Output: Return 1, and the first 1 characters of the input array should be: ["a"] Explanation: Nothing is replaced.
Example 3:
Input: ["a","b","b","b","b","b","b","b","b","b","b","b","b"] Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"]. Explanation: Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12". Notice each digit has it's own entry in the array.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
在数组中输入一个1 2时,需要start i同时变
[一句话思路]:
往前走多长,然后截断、更新,属于 前向窗口类:end边统计往后走到哪,start边统计可以更新到哪
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
-
String.valueOf 函数可以将数字转化成字符串。不是.toString,这是针对已有的字符串
- 只有count != 1时才需要添加,读题时就要备注特殊条件
[二刷]:
- 用等号表示的最末尾一位是n - 1,也需要备注
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
字符型数组应写成char[], 不是chars[]
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(1)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
for 循环+ j 满足条件时移动
for (int end = 0, count = 0; end < chars.length; end++) { count++; //change if neccessary if (end == chars.length - 1 || chars[end] != chars[end + 1]) { chars[start] = chars[end]; start++;
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
Encode and Decode Strings 用各种类的设计,有基础就还行
[代码风格] :
写框架时只能把for的重要事项写好,具体换行还是需要自己理解
class Solution { public int compress(char[] chars) { //cc if (chars.length == 0) { return 0; } //ini int start = 0; for (int end = 0, count = 0; end < chars.length; end++) { count++; //change if neccessary if (end == chars.length - 1 || chars[end] != chars[end + 1]) { chars[start] = chars[end]; start++; //add to only if (count != 1) if (count != 1) { char[] arrs = String.valueOf(count).toCharArray(); for (int i = 0; i < arrs.length; i++, start++) { chars[start] = arrs[i]; } } //reset count count = 0; } } //return start; return start; } }