Leetcode 38 - Count and say

Note:

  • Update:
    • Use one variable to store prev result.
    • Using stack is more wise.
  • The description of this question is quite confusing.
  • Even it’s a medium question, it’s not that hard. Don’t be terrified!

Question:

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

  • countAndSay(1) = “1”
  • countAndSay(n) is the way you would “say” the digit string from countAndSay(n-1), which is then converted into a different digit string.

To determine how you “say” a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

For example, the saying and conversion for digit string “3322251“:
img
Given a positive integer n, return the nth term of the count-and-say sequence.

Example:

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Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"

Code:

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/**
* @param {number} n
* @return {string}
*/
var countAndSay = function (n) {
let curr = '1';
let stack = [];
for (let i = 2; i <= n; i++) {
let tmp = '';
for (const char of curr) {
if (char === stack[stack.length - 1]) {
stack.push(char);
} else {
if (stack.length > 0) {
tmp += stack.length + '' + stack[0];
stack = [];
stack.push(char);
} else {
stack.push(char);
}
}
}
curr = tmp;
curr += stack.length + '' + stack[0];
stack = [];
}
return curr;
};