443. String Compression 字符串壓縮
阿新 • • 發佈:2017-11-04
you 2.6 bottom none var lac ++ enum sin
來自為知筆記(Wiz)
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
.
/**
* @param {character[]} chars
* @return {number}
*/
var compress = function(chars) {
if(!chars){
return 0
}
if(chars.length <= 1){
return chars.length;
}
let res = "";
let count = chars.length - 1;
let lastChar = chars[0];
let repeat = 0;
for(let i = 1; i <= count; i++){
let curChar = chars[i];
if(curChar == lastChar){
repeat++;
}
if(curChar != lastChar || i == count){
repeat <= 0 ? (res += lastChar) : (res += lastChar + String(repeat+1));
repeat = 0;
lastChar = curChar;
}
}
if(chars[chars.length-1] != chars[chars.length-2]){
res+=chars[chars.length-1];
}
res = res.split("");
for(let i in res){
chars[i] = res[i];
}
chars.length = res.length;
return res.length;
};
來自為知筆記(Wiz)
443. String Compression 字符串壓縮