Category: 技術

還滿難的一道題。

題目描述：

編寫一個函數來查找字符串數組中最長的共有前綴字符串。

如果沒有共有前綴，則返回一個空字符串""。

Example 1:

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

Example 2:

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

Horizontal scanning:

自己嘗試的方法：

class Solution {
    public String longestCommonPrefix(String[] strs)
    {
        int size = strs.length;
        if (size == 0)
            return "";

        if (size == 1)
            return strs[0];

        Arrays.sort(strs);
        int end = Math.min(strs[0].length(), strs[size-1].length());
        int i = 0;
        while (i < end && strs[0].charAt(i) == strs[size-1].charAt(i) )
            i++;
        String pre = strs[0].substring(0, i);
        return pre;
    }
}

1 ms

Vertical scanning:

Solution 中提供的方法二：

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    for (int i = 0; i < strs[0].length() ; i++){
        char c = strs[0].charAt(i);
        for (int j = 1; j < strs.length; j ++) {
            if (i == strs[j].length() || strs[j].charAt(i) != c)
                return strs[0].substring(0, i);             
        }
    }
    return strs[0];
}

Divide and conquer:

Solution 中提供的方法三：

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";    
        return longestCommonPrefix(strs, 0 , strs.length - 1);
}

private String longestCommonPrefix(String[] strs, int l, int r) {
    if (l == r) {
        return strs[l];
    }
    else {
        int mid = (l + r)/2;
        String lcpLeft =   longestCommonPrefix(strs, l , mid);
        String lcpRight =  longestCommonPrefix(strs, mid + 1,r);
        return commonPrefix(lcpLeft, lcpRight);
   }
}

String commonPrefix(String left,String right) {
    int min = Math.min(left.length(), right.length());       
    for (int i = 0; i < min; i++) {
        if ( left.charAt(i) != right.charAt(i) )
            return left.substring(0, i);
    }
    return left.substring(0, min);
}

Binary search:

Solution 中提供的方法四：

public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0)
        return "";
    int minLen = Integer.MAX_VALUE;
    for (String str : strs)
        minLen = Math.min(minLen, str.length());
    int low = 1;
    int high = minLen;
    while (low <= high) {
        int middle = (low + high) / 2;
        if (isCommonPrefix(strs, middle))
            low = middle + 1;
        else
            high = middle - 1;
    }
    return strs[0].substring(0, (low + high) / 2);
}

private boolean isCommonPrefix(String[] strs, int len){
    String str1 = strs[0].substring(0,len);
    for (int i = 1; i < strs.length; i++)
        if (!strs[i].startsWith(str1))
            return false;
    return true;
}

題目描述：

將羅馬數字轉換為整數，具體規則略。

Example 1:

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

Example 2:

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

Example 3:

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

自己的 Switch 暴力解法：

兜了好幾個圈……

class Solution {
    public int romanToInt(String s) {
        int num = 0;
        StringBuilder re = new StringBuilder();
        re.append(s).reverse();
        String[] split = re.toString().split("(?<=\\G..)");
        for (String i:split){
            switch(i){
                case "VI":
                    num = num + 4;
                    break;
                case "XI":
                    num = num + 9;
                    break;
                case "LX":
                    num = num + 40; 
                    break;
                case "CX":
                    num = num + 90; 
                    break;
                case "DC":
                    num = num + 400;
                    break;
                case "MC":
                    num = num + 900; 
                    break;
                default:
                    String[] sub = i.split("");
                    for (String j:sub){
                        switch(j){
                            case "I":
                                num = num + 1;
                                break;
                            case "V":
                                num = num + 5;
                                break;
                            case "X":
                                num = num + 10; 
                                break;
                            case "L":
                                num = num + 50; 
                                break;
                            case "C":
                                num = num + 100;
                                break;
                            case "D":
                                num = num + 500; 
                                break;
                            case "M":
                                num = num + 1000;     
                                break;
                        }
                    }
                }
        }
        return num;    
    }
}

意外發生了，本地 IDE 能跑出正確答案，但是在 Leetcode 上跑不出來，我竟然為這種問題浪費了不少時間，最終也沒解決……

正確精練的 Switch 解法：

class Solution {
	 public int romanToInt(String s) {
	    int nums[] = new int[s.length()];
	    for(int i = 0; i < s.length(); i++){
	        switch (s.charAt(i)) {
	            case 'M':
	                nums[i] = 1000;
	                break;
	            case 'D':
	                nums[i] = 500;
	                break;
	            case 'C':
	                nums[i] = 100;
	                break;
	            case 'L':
	                nums[i] = 50;
	                break;
	            case 'X' :
	                nums[i] = 10;
	                break;
	            case 'V':
	                nums[i] = 5;
	                break;
	            case 'I':
	                nums[i] = 1;
	                break;
	        }
	    } 
	    
	    int sum = 0;
	    for(int i=0; i<nums.length-1; i++){
	        if(nums[i] < nums[i+1])
	            sum -= nums[i];
	        else
	            sum += nums[i];
	    }
         
	    return sum + nums[nums.length-1];
	}
}

6 ms

運用 Hashmap：

class solution {
    public int romanToInt(String s) {
        Map map = new HashMap<>();
        map.put('I',1);
        map.put('V',5);
        map.put('X',10);
        map.put('L',50);
        map.put('C',100);
        map.put('D',500);
        map.put('M',1000);
        
        int result = map.get(s.charAt(s.length()-1));
        for(int i=s.length()-2; i>=0; i--){
            if(map.get(s.charAt(i)) < map.get(s.charAt(i+1))){
                result -= map.get(s.charAt(i));
            }
            else{
                result += map.get(s.charAt(i));
            }
        }
        return result;
    }
}

題目描述：

給定一個整數 x，如果 x 是回文數，則返回 true。

回文數指一個整數向後讀和向前讀内容相同。

Example 1:

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

Example 2:

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:

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

算術解法：

利用 Java 整數倒序計算：

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        
        int ori = x;
        int p = 0;
        while (ori != 0) {
            p = p * 10 + ori % 10;
            ori /= 10;
        }
        
        return x == p;
    }
}

9 ms，超過 87.67% 的答案。

轉換為 String 的解法：

進階要求 Follow Up 竟然問了一句 Could you solve it without converting the integer to a string? 原來我想到的已經是進階方法了，常規解法是轉換為 String：

class Solution {
     public boolean isPalindrome(int x) {
         char[] nums = String.valueOf(x).toCharArray();
         int start = 0;
         int end = nums.length-1;
         while(start < end) {
             if(nums[start] != nums[end]) return false;
             start++; end--;
         }
        return true;
    }
}

19 ms，超過 19.48% 的答案。