LeetCode on Yen's 實作筆記

LeetCode 實作解說系列-Longest Common Prefix

Sun, 19 Jun 2022 00:00:00 +0000

題目解說

原題目

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

1
2

Input: strs = ["flower","flow","flight"]
Output: "fl"

Example 2:

1
2
3

Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

Constraints:

1 <= strs.length <= 200
0 <= strs[i].length <= 200
strs[i] consists of only lowercase English letters.

翻譯

給定一個陣列strs，裡面會有字串，找出此陣列字串相同的前綴，如果沒有，返回""。

解題策略

邏輯上是後面的字串去跟前面的做比較就好，所以聲明字串pre，一開始就宣告為陣列的第一個元素，如果比對不到，就將pre刪掉一個字符，刪到比對出相同前綴為止。

JS程式碼

var longestCommonPrefix = function(strs) {
   if(strs == null || strs.length == 0) return "";
   let pre =strs[0]
   console.log(strs.length)
   for(let i =1;i<strs.length;i++){
       while(strs[i].indexOf(pre) != 0){
           pre=pre.substring(0, pre.length - 1);
       }
   }
    return pre;
};

LeetCode 實作解說系列-Roman to Integer

Sun, 19 Jun 2022 00:00:00 +0000

題目解說

原題目

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer.

Example 1:

1
2
3

Input: s = "III"
Output: 3
Explanation: III = 3.

Example 2:

1
2
3

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 3:

1
2
3

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
It is guaranteed that s is a valid roman numeral in the range [1, 3999].

翻譯

這題主要是在說羅馬數字的計算方式，並告知('I', 'V', 'X', 'L', 'C', 'D', 'M') 分別等於數字多少，且 4 不是 IIII 而是 VI ，以此類推應用到 9,40,90,400,900。

解題策略

解題邏輯是讓每個字母等於一個數字，再用 object ={key:value} 的方式加總就好，首先要處理的就是 4,9,40,90,400,900 ，我宣告的 object 把 'IV':4,'IX':9,'XL':40,'XC':90,'CD':400,'CM':900 用一個字母取代掉 'f':4,'n':9,'l':40,'c':90,'d':400,'m':900，接著是做字串s的文字取代 s.replace('IV','f').replace('IX','n').replace('XL','l').replace('XC','c').replace('CD','d').replace('CM','m')，最後利用迴圈加上 charAt()就可以逐一取值並加總。

JS程式碼

var romanToInt = function(s) {
    //'IV':4,'IX':9,'XL':40,'XC':90,'CD':400,'CM':900
    const obj  ={'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000,'f':4,'n':9,'l':40,'c':90,'d':400,'m':900};
    let new_s = s.replace('IV','f').replace('IX','n').replace('XL','l').replace('XC','c').replace('CD','d').replace('CM','m')
    let sum = 0
    for(let i = 1;i<=new_s.length;i++){
       sum = sum + obj[new_s.charAt(i-1)]
    }
    return sum;
};

LeetCode 實作解說系列-Palindrome Number

Sat, 18 Jun 2022 00:00:00 +0000

題目解說

原題目

Given an integer x, return true if x is palindrome integer.

An integer is a palindrome when it reads the same backward as forward.

For example, 121 is a palindrome while 123 is not.

Example 1:

1
2
3

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left

Example 2:

1
2
3

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

1
2
3

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Constraints:

-231 <= x <= 231 - 1

翻譯

給定一個整數x，回傳x是不是個相反數。舉例來說121相反為121，回傳true，-121相反為121-，回傳false。

解題策略

將x轉為字串xstr，再將xstr利用reverse函數做字序的反轉，最後將x與xstr做一般相等比較。

JS程式碼

var isPalindrome = function(x) {
    let xstr = x.toString().split("").reverse().join("")
    console.log(xstr)
    if(x==xstr){
        return true;
    }
    else{
        return false;
    }
};

LeetCode 實作解說系列-Two Sum

Sat, 18 Jun 2022 00:00:00 +0000

題目解說

原題目

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

翻譯

給一個整數陣列nums以及一個整數target，回傳出兩個整數相加等於target在nums裡面的索引值，每個數不能被重複使用，而且只會有一個解。

Example 1:

1
2
3

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

1
2

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

1
2

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.

解題策略

跑過整個陣列後必定能找到兩兩相加為target的值，用雙層迴圈跑完陣列中的所有數並兩兩相加，方得解答。

JS程式碼

var twoSum = function(nums, target) {
    for(let i = 0;i<nums.length-1;i++){
        let x = nums[i];
        for(let j = i+1;j<nums.length;j++)
            {
            let  y = nums[j];
            if(x+y===target){
                return [i,j]
            }
            }
 
    }
};