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English Version

题目描述

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。

 

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

解法

Python3

class Solution:
    def restoreString(self, s: str, indices: List[int]) -> str:
        ans = [0] * len(s)
        for i, c in enumerate(s):
            ans[indices[i]] = c
        return ''.join(ans)

Java

class Solution {
    public String restoreString(String s, int[] indices) {
        int n = s.length();
        char[] ans = new char[n];
        for (int i = 0; i < n; ++i) {
            ans[indices[i]] = s.charAt(i);
        }
        return String.valueOf(ans);
    }
}

C++

class Solution {
public:
    string restoreString(string s, vector<int>& indices) {
        int n = s.size();
        string ans(n, 0);
        for (int i = 0; i < n; ++i) {
            ans[indices[i]] = s[i];
        }
        return ans;
    }
};

Go

func restoreString(s string, indices []int) string {
	ans := make([]rune, len(s))
	for i, c := range s {
		ans[indices[i]] = c
	}
	return string(ans)
}

...