Question
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
.
Analyze
这道题目是计算排序好的字母的每个重复个数,然后转换成其他方式输出出来,如果超过1个的字母就输入字母+个数。处理方式如下:
- 定义两个指针
left=0
和right=0
, 移动right
指针 - 如果
right
指针指向的元素等于指针left
指向的元素,那么就继续移动指针right
- 如果
right
指针指向的元素不等于指针left
指向的元素,那么就将指针left
的值赋值给最开始的一个元素,然后left
自加1;如果此时right-left ==1
,那么就跳出这一次循环,如果没有,那么就将right-left
赋值给char[cur++]
实现将大约1的元素个数存到元素字母的后面。 - 返回
cur
Solution
class Solution {
public:
int compress(vector<char>& chars) {
int n = chars.size(), cur = 0;
for (int left = 0, right = 0; left < n; left = right) {
while (right < n && chars[right] == chars[left]) ++right;
chars[cur++] = chars[left];
if (right - left == 1) continue;
for (char c : to_string(right - left)) chars[cur++] = c;
}
return cur;
}
};