难度:Easy
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.
原地修改,需要注意重复元素出现次数,如果是1可以省略,后面补上的长度是次数的宽度。
class Solution {public:int compress(vector<char>& chars) {int start=1;int count=1;int res=0;for(int i=1;i<chars.size();i++){if(chars[i]==chars[i-1])count++;else{if(count==1){res+=1;}else{string tmp=to_string(count);res+=1+tmp.length();for(auto c:tmp)chars[start++]=c;count=1;}chars[start++]=chars[i];}}if(count==1){res+=1;}else{string tmp=to_string(count);res+=1+tmp.length();for(auto c:tmp)chars[start++]=c;}return res;}};
执行用时 :12 ms, 在所有 C++ 提交中击败了89.58%的用户 内存消耗 :9.1 MB, 在所有 C++ 提交中击败了91.06%的用户