add: 只出现一次的数字等

src/string/restore-ip-addresses.js

@@ -0,0 +1,56 @@
+// LeetCode 93. 复原IP地址 https://leetcode-cn.com/problems/restore-ip-addresses/
+// LintCode 426. 恢复IP地址 https://www.lintcode.com/problem/restore-ip-addresses/description
+
+export default (str) => {
+  const result = []
+
+  if (str === '0000') return ['0.0.0.0']
+
+  // 递归函数
+  const recur = (cur, sub) => {
+    if (cur.length === 4 && cur.join('') === str) {
+      if (!cur.every(item => Number(item) === 0)) {
+        result.push(cur.join('.'))
+      }
+    } else {
+      for (let n = 0, len = Math.min(3, sub.length); n < len; n++) {
+        const start = Number(sub.substr(0, n + 1))
+        const end = sub.substr(n + 1)
+        if (start < 256 && end.length <= (9 - cur.length * 3)) {
+          recur([...cur, start], end)
+        }
+      }
+    }
+  }
+
+  recur([], str)
+  return result
+}
+
+// 快乐动起来老师的解法：
+// export default (str) => {
+//   // 保存所有符合条件的IP地址
+//   let r = []
+//   // 分四步递归处理ip分段
+//   let search = (cur, sub) => {
+//     // 非法输入过滤，LeetCode测试用例(111111111111111111111111111111111111111111111111111111111111)
+//     if (sub.length > 12) {
+//       return
+//     }
+//     // 边界条件
+//     if (cur.length === 4 && cur.join('') === str) {
+//       r.push(cur.join('.'))
+//     } else {
+//       // 正常的处理过程
+//       for (let i = 0, len = Math.min(3, sub.length), tmp; i < len; i++) {
+//         tmp = sub.substr(0, i + 1)
+//         if (tmp - 256 < 0) {
+//           // 转换下数据类型，如 01为1（LeetCode测试用例）
+//           search(cur.concat([tmp * 1]), sub.substr(i + 1))
+//         }
+//       }
+//     }
+//   }
+//   search([], str)
+//   return r
+// }