mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 08:27:30 +08:00
99 lines
2.1 KiB
Markdown
99 lines
2.1 KiB
Markdown
# [821. Shortest Distance to a Character](https://leetcode.com/problems/shortest-distance-to-a-character/)
|
||
|
||
## 题目
|
||
|
||
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 shortest distance from* `s[i]` *to the character* `c` *in* `s`.
|
||
|
||
**Example 1:**
|
||
|
||
```
|
||
Input: s = "loveleetcode", c = "e"
|
||
Output: [3,2,1,0,1,0,0,1,2,2,1,0]
|
||
```
|
||
|
||
**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.
|
||
- `c` occurs at least once in `s`.
|
||
|
||
## 题目大意
|
||
|
||
给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。
|
||
|
||
## 解题思路
|
||
|
||
- 解法一：从左至右更新一遍到 C 的值距离，再从右至左更新一遍到 C 的值，取两者中的最小值。
|
||
- 解法二：依次扫描字符串 S，针对每一个非字符 C 的字符，分别往左扫一次，往右扫一次，计算出距离目标字符 C 的距离，然后取左右两个距离的最小值存入最终答案数组中。
|
||
|
||
## 代码
|
||
|
||
```go
|
||
package leetcode
|
||
|
||
import (
|
||
"math"
|
||
)
|
||
|
||
// 解法一
|
||
func shortestToChar(s string, c byte) []int {
|
||
n := len(s)
|
||
res := make([]int, n)
|
||
for i := range res {
|
||
res[i] = n
|
||
}
|
||
for i := 0; i < n; i++ {
|
||
if s[i] == c {
|
||
res[i] = 0
|
||
} else if i > 0 {
|
||
res[i] = res[i-1] + 1
|
||
}
|
||
}
|
||
for i := n - 1; i >= 0; i-- {
|
||
if i < n-1 && res[i+1]+1 < res[i] {
|
||
res[i] = res[i+1] + 1
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
|
||
// 解法二
|
||
func shortestToChar1(s string, c byte) []int {
|
||
res := make([]int, len(s))
|
||
for i := 0; i < len(s); i++ {
|
||
if s[i] == c {
|
||
res[i] = 0
|
||
} else {
|
||
left, right := math.MaxInt32, math.MaxInt32
|
||
for j := i + 1; j < len(s); j++ {
|
||
if s[j] == c {
|
||
right = j - i
|
||
break
|
||
}
|
||
}
|
||
for k := i - 1; k >= 0; k-- {
|
||
if s[k] == c {
|
||
left = i - k
|
||
break
|
||
}
|
||
}
|
||
res[i] = min(left, right)
|
||
}
|
||
}
|
||
return res
|
||
}
|
||
|
||
func min(a, b int) int {
|
||
if a > b {
|
||
return b
|
||
}
|
||
return a
|
||
}
|
||
``` |