标签：compress turn flow var padding ott character pac let
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:
[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;};
标签：compress turn flow var padding ott character pac let
原文地址：http://www.cnblogs.com/xiejunzhao/p/7784087.html
