443. String Compression
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问题描述:
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.
Note:
- All characters have an ASCII value in
[35, 126]
. 1 <= len(chars) <= 1000
.
解题思路:
在现有位置上压缩,用cur来记录当前字符,count记录出现个数,若遍历到的字符:
1. chars[i] == cur : count++;
2.chars[i] != cur:将cur及其出现次数更新在chars中, 需要注意的是:当cur>1时,才把数字记入字符数组中
另有一点需要注意:
推出循环时,仍有最后一个字符没有计入到数组中!所以需再加一遍。
代码:
class Solution { public: int compress(vector<char>& chars) { int n = chars.size(); if(n < 2) return n; char cur = chars[0]; int count = 0; int pos = 0; for(int i = 0; i < n; i++){ if(chars[i] == cur){ count++; }else{ chars[pos] = cur; pos++; if(count > 1){ string digit = to_string(count); for(int j = 0; j < digit.size(); j++) chars[pos++] = digit[j]; } count = 1; cur = chars[i]; } } chars[pos++] = cur; if(count > 1){ string digit = to_string(count); for(int j = 0; j < digit.size(); j++) chars[pos++] = digit[j]; } return pos; } };
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leetcode443. String Compression