Shifting Letters

最新推荐文章于 2025-05-31 01:15:00 发布

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最新推荐文章于 2025-05-31 01:15:00 发布

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本文深入探讨了一种处理字符串位移的算法，该算法通过应用一系列整数位移，对字符串中的字母进行重新排列。具体而言，对于输入的字符串S和整数数组shifts，算法会将第一个i+1个字母按shifts[i]的次数进行位移，最终返回经过所有位移操作后的字符串。文章还提供了一个示例，展示了如何从abc开始，通过三次位移操作得到rpl。

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We have a string S of lowercase letters, and an integer array shifts.

Call the shift of a letter, the next letter in the alphabet, (wrapping around so that ‘z’ becomes ‘a’).

For example, shift(‘a’) = ‘b’, shift(‘t’) = ‘u’, and shift(‘z’) = ‘a’.

Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.

Return the final string after all such shifts to S are applied.

Example 1:

Input: S = “abc”, shifts = [3,5,9]
Output: “rpl”
Explanation:
We start with “abc”.
After shifting the first 1 letters of S by 3, we have “dbc”.
After shifting the first 2 letters of S by 5, we have “igc”.
After shifting the first 3 letters of S by 9, we have “rpl”, the answer.
Note:

1 <= S.length = shifts.length <= 20000
0 <= shifts[i] <= 10 ^ 9

class Solution {
    public String shiftingLetters(String S, int[] shifts) {
          if(S==null||S.isEmpty())
            return S;
        StringBuilder sb=new StringBuilder(S);
        for(int i=0;i<sb.length();i++){
            for(int j=0;j<=i;j++){
               sb.setCharAt(j,(char)(((sb.charAt(j)-'a'+shifts[i])%26+'a')));
                
            }
        }
        return sb.toString();
    }
}

这道题使用StringBuffer时时间超过限制Time Limit Exceeded,使用StringBuilder才能通过，通过的思路是在遍历shifts的时候，将它加到i之前的所有已经遍历过的字符上。