string-compression

String Compression – LeetCode 443

Problem

Description

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?

Answer

Original

Code

class Solution {
public:
    int compress(vector<char>& chars) {
        if(!chars.size()) return 0;
        char c = ' ';
        int cnt = 0;
        auto wh = chars.begin();
        for(auto ch = chars.begin(); ch != chars.end(); ++ch){
            if(c == ' '){
                c = *ch;
                ++cnt;
            }else if(c != *ch){
                *wh++ = c;
                if(cnt != 1) {
                    for(auto &c: to_string(cnt))
                        *wh++ = c;
                }
                c = *ch;
                cnt = 1;
            }else{
                ++cnt;
            }
        }
        *wh++ = c;
        if(cnt != 1) {
            for(auto &c: to_string(cnt))
                *wh++ = c;
        }
        chars.erase(wh,chars.end());
        return chars.size();
    }
};

思路

简单的遍历后完成压缩。时间复杂度$O(n)$，空间复杂度$O(1)$。
耗时$8$ ms，排名$72.79\%$

Better

思路

思路相同，写法更简洁。时间复杂度$O(n)$，空间复杂度$O(1)$。
耗时$6$ ms，排名$72.73\%$

Code

class Solution {
public:
    int compress(vector<char>& chars) {
        int lo=0;
        int cnt=0;
        for(int i=0; i<chars.size(); i++){
            cnt++;
            if(i==chars.size()-1||chars[i]!=chars[i+1]){
                chars[lo++]=chars[i];
                if(cnt>1){
                    string nums=to_string(cnt);
                    for(int i=0; i<nums.length(); i++){
                        chars[lo++]=nums[i];
                    }
                }
                cnt=0;
            }
        }
        return lo;
    }
};