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1528-重新排列字符串(Shuffle String)
发表于:2021-12-03 | 分类: 简单
字数统计: 601 | 阅读时长: 3分钟 | 阅读量:

原文链接: https://leetcode-cn.com/problems/shuffle-string

英文原文

Given a string s and an integer array indices of the same length.

The string s will be shuffled such that the character at the ith position moves to indices[i] in the shuffled string.

Return the shuffled string.

 

Example 1:

Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.

Example 2:

Input: s = "abc", indices = [0,1,2]
Output: "abc"
Explanation: After shuffling, each character remains in its position.

Example 3:

Input: s = "aiohn", indices = [3,1,4,2,0]
Output: "nihao"

Example 4:

Input: s = "aaiougrt", indices = [4,0,2,6,7,3,1,5]
Output: "arigatou"

Example 5:

Input: s = "art", indices = [1,0,2]
Output: "rat"

 

Constraints:

  • s.length == indices.length == n
  • 1 <= n <= 100
  • s contains only lower-case English letters.
  • 0 <= indices[i] < n
  • All values of indices are unique (i.e. indices is a permutation of the integers from 0 to n - 1).

中文题目

给你一个字符串 s 和一个 长度相同 的整数数组 indices

请你重新排列字符串 s ,其中第 i 个字符需要移动到 indices[i] 指示的位置。

返回重新排列后的字符串。

 

示例 1:

输入:s = "codeleet", indices = [4,5,6,7,0,2,1,3]
输出:"leetcode"
解释:如图所示,"codeleet" 重新排列后变为 "leetcode" 。

示例 2:

输入:s = "abc", indices = [0,1,2]
输出:"abc"
解释:重新排列后,每个字符都还留在原来的位置上。

示例 3:

输入:s = "aiohn", indices = [3,1,4,2,0]
输出:"nihao"

示例 4:

输入:s = "aaiougrt", indices = [4,0,2,6,7,3,1,5]
输出:"arigatou"

示例 5:

输入:s = "art", indices = [1,0,2]
输出:"rat"

 

提示:

  • s.length == indices.length == n
  • 1 <= n <= 100
  • s 仅包含小写英文字母。
  • 0 <= indices[i] < n
  • indices 的所有的值都是唯一的(也就是说,indices 是整数 0n - 1 形成的一组排列)。

通过代码

高赞题解

解题思路

此处撰写解题思路

代码

class Solution {
    public String restoreString(String s, int[] indices) {
             TreeMap<Integer,String> map=new TreeMap<>();
            for(int i=0;i<indices.length;i++){
                map.put(indices[i],s.charAt(i)+"");
            }
            StringBuffer stringBuffer=new StringBuffer();
            for(int i:map.keySet()){
                stringBuffer.append(map.get(i));
            }
            return new String(stringBuffer);
    }
}

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