Merge pull request #124 from NovaHe/feature/821

feature/821: add O(n) solution

Author: halfrost

leetcode/0821.Shortest-Distance-to-a-Character/821. Shortest Distance to a Character.go

@@ -4,7 +4,30 @@ 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 {

leetcode/0821.Shortest-Distance-to-a-Character/README.md

@@ -1,6 +1,5 @@
 # [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`.
@@ -31,7 +30,8 @@ Output: [3,2,1,0]
 
 ## 解题思路
 
-- 简单题。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。
+- 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
 
@@ -42,7 +42,30 @@ 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 {