821. 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

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
}

Uh oh!