Question

942. 增减字符串匹配

English Version

题目描述

由范围 [0,n] 内所有整数组成的 n + 1 个整数的排列序列可以表示为长度为 n 的字符串 s ，其中:

• 如果 perm[i] < perm[i + 1] ，那么 s[i] == 'I' 
• 如果 perm[i] > perm[i + 1] ，那么 s[i] == 'D' 

给定一个字符串 s ，重构排列 perm 并返回它。如果有多个有效排列perm，则返回其中 任何一个 。

 

示例 1：

输入：s = "IDID"
输出：[0,4,1,3,2]

示例 2：

输入：s = "III"
输出：[0,1,2,3]

示例 3：

输入：s = "DDI"
输出：[3,2,0,1]

 

提示：

• 1 <= s.length <= 105
• s 只包含字符 "I" 或 "D"

解法

方法一

class Solution:
  def diStringMatch(self, s: str) -> List[int]:
    n = len(s)
    low, high = 0, n
    ans = []
    for i in range(n):
      if s[i] == 'I':
        ans.append(low)
        low += 1
      else:
        ans.append(high)
        high -= 1
    ans.append(low)
    return ans

class Solution {
  public int[] diStringMatch(String s) {
    int n = s.length();
    int low = 0, high = n;
    int[] ans = new int[n + 1];
    for (int i = 0; i < n; i++) {
      if (s.charAt(i) == 'I') {
        ans[i] = low++;
      } else {
        ans[i] = high--;
      }
    }
    ans[n] = low;
    return ans;
  }
}

class Solution {
public:
  vector<int> diStringMatch(string s) {
    int n = s.size();
    int low = 0, high = n;
    vector<int> ans(n + 1);
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'I') {
        ans[i] = low++;
      } else {
        ans[i] = high--;
      }
    }
    ans[n] = low;
    return ans;
  }
};

func diStringMatch(s string) []int {
  n := len(s)
  low, high := 0, n
  var ans []int
  for i := 0; i < n; i++ {
    if s[i] == 'I' {
      ans = append(ans, low)
      low++
    } else {
      ans = append(ans, high)
      high--
    }
  }
  ans = append(ans, low)
  return ans
}

function diStringMatch(s: string): number[] {
  const n = s.length;
  const res = new Array(n + 1);
  let low = 0;
  let high = n;
  for (let i = 0; i < n; i++) {
    if (s[i] === 'I') {
      res[i] = low++;
    } else {
      res[i] = high--;
    }
  }
  res[n] = low;
  return res;
}

impl Solution {
  pub fn di_string_match(s: String) -> Vec<i32> {
    let s = s.as_bytes();
    let n = s.len();
    let mut res = Vec::with_capacity(n + 1);
    let (mut low, mut high) = (-1, (n + 1) as i32);
    for i in 0..n {
      res.push(
        if s[i] == b'I' {
          low += 1;
          low
        } else {
          high -= 1;
          high
        }
      );
    }
    res.push(low + 1);
    res
  }
}