add: 左旋转字符串

.vscode/settings.json

@@ -1,11 +1,18 @@
 {
   "cSpell.words": [
+    "abcdefg",
+    "cdefgab",
+    "chuan",
     "dvdf",
     "lcci",
     "lcof",
+    "lrloseumgh",
     "mincost",
     "nums",
     "pwwkew",
-    "zhong"
+    "umghlrlose",
+    "xuan",
+    "zhong",
+    "zhuan"
   ]
 }

README.md

@@ -67,6 +67,10 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
   - 面试题参考思路，不严谨实现 廖雪峰 不要使用JavaScript内置的parseInt()函数，利用map和reduce操作实现一个string2int()函数。 https://www.liaoxuefeng.com/wiki/1022910821149312/1024322552460832
 
+- [左旋转字符串](src/string/zuo-xuan-zhuan-zi-fu-chuan-lcof.js)
+
+  - LeetCode 面试题58 - II. 左旋转字符串 https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/
+
 ## 数组
 
 - [电话号码的字母组合](src/array/letter-combinations-of-a-phone-number.js)

src/string/zuo-xuan-zhuan-zi-fu-chuan-lcof.js

@@ -0,0 +1,8 @@
+/**
+ * @param {string} s
+ * @param {number} n
+ * @return {string}
+ */
+export const reverseLeftWords = function (s, n) {
+  return (s + s).substr(n, s.length)
+}

test/string/zuo-xuan-zhuan-zi-fu-chuan-lcof.test.js

@@ -0,0 +1,6 @@
+import { reverseLeftWords } from '../../src/string/zuo-xuan-zhuan-zi-fu-chuan-lcof'
+
+test('左旋转字符串', () => {
+  expect(reverseLeftWords('abcdefg', 2)).toBe('cdefgab')
+  expect(reverseLeftWords('lrloseumgh', 6)).toBe('umghlrlose')
+})