Question

821. Shortest Distance to a Character

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Description

Given a string s and a character c that occurs in s, return _an array of integers _answer_ where _answer.length == s.length_ and _answer[i]_ is the distance from index _i_ to the closest occurrence of character _c_ in _s.

The distance between two indices i and j is abs(i - j), where abs is the absolute value function.

 

Example 1:

Input: s = "loveleetcode", c = "e"
Output: [3,2,1,0,1,0,0,1,2,2,1,0]
Explanation: The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.

Example 2:

Input: s = "aaab", c = "b"
Output: [3,2,1,0]

 

Constraints:

• 1 <= s.length <= 104
• s[i] and c are lowercase English letters.
• It is guaranteed that c occurs at least once in s.

Solutions

Solution 1

class Solution:
  def shortestToChar(self, s: str, c: str) -> List[int]:
    n = len(s)
    ans = [n] * n
    pre = -inf
    for i, ch in enumerate(s):
      if ch == c:
        pre = i
      ans[i] = min(ans[i], i - pre)
    suf = inf
    for i in range(n - 1, -1, -1):
      if s[i] == c:
        suf = i
      ans[i] = min(ans[i], suf - i)
    return ans

class Solution {
  public int[] shortestToChar(String s, char c) {
    int n = s.length();
    int[] ans = new int[n];
    final int inf = 1 << 30;
    Arrays.fill(ans, inf);
    for (int i = 0, pre = -inf; i < n; ++i) {
      if (s.charAt(i) == c) {
        pre = i;
      }
      ans[i] = Math.min(ans[i], i - pre);
    }
    for (int i = n - 1, suf = inf; i >= 0; --i) {
      if (s.charAt(i) == c) {
        suf = i;
      }
      ans[i] = Math.min(ans[i], suf - i);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> shortestToChar(string s, char c) {
    int n = s.size();
    const int inf = 1 << 30;
    vector<int> ans(n, inf);
    for (int i = 0, pre = -inf; i < n; ++i) {
      if (s[i] == c) {
        pre = i;
      }
      ans[i] = min(ans[i], i - pre);
    }
    for (int i = n - 1, suf = inf; ~i; --i) {
      if (s[i] == c) {
        suf = i;
      }
      ans[i] = min(ans[i], suf - i);
    }
    return ans;
  }
};

func shortestToChar(s string, c byte) []int {
  n := len(s)
  ans := make([]int, n)
  const inf int = 1 << 30
  pre := -inf
  for i := range s {
    if s[i] == c {
      pre = i
    }
    ans[i] = i - pre
  }
  suf := inf
  for i := n - 1; i >= 0; i-- {
    if s[i] == c {
      suf = i
    }
    ans[i] = min(ans[i], suf-i)
  }
  return ans
}

function shortestToChar(s: string, c: string): number[] {
  const n = s.length;
  const inf = 1 << 30;
  const ans: number[] = new Array(n).fill(inf);
  for (let i = 0, pre = -inf; i < n; ++i) {
    if (s[i] === c) {
      pre = i;
    }
    ans[i] = i - pre;
  }
  for (let i = n - 1, suf = inf; i >= 0; --i) {
    if (s[i] === c) {
      suf = i;
    }
    ans[i] = Math.min(ans[i], suf - i);
  }
  return ans;
}

impl Solution {
  pub fn shortest_to_char(s: String, c: char) -> Vec<i32> {
    let c = c as u8;
    let s = s.as_bytes();
    let n = s.len();
    let mut res = vec![i32::MAX; n];
    let mut pre = i32::MAX;
    for i in 0..n {
      if s[i] == c {
        pre = i as i32;
      }
      res[i] = i32::abs((i as i32) - pre);
    }
    pre = i32::MAX;
    for i in (0..n).rev() {
      if s[i] == c {
        pre = i as i32;
      }
      res[i] = res[i].min(i32::abs((i as i32) - pre));
    }
    res
  }
}