添加 problem 125

Algorithms/125.Valid-Palindrome/README.md

@@ -0,0 +1,18 @@
+# [125. Valid Palindrome](https://leetcode.com/problems/valid-palindrome/description/)
+
+## 题目
+
+Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+For example,
+"A man, a plan, a canal: Panama" is a palindrome.
+"race a car" is not a palindrome.
+
+Note:
+Have you consider that the string might be empty? This is a good question to ask during an interview.
+
+For the purpose of this problem, we define empty string as valid palindrome.
+
+## 题目大意
+
+判断所给的字符串是否是有效的回文串

Algorithms/125.Valid-Palindrome/Valid Palindrome.go

@@ -0,0 +1,35 @@
+package leetcode
+
+import (
+  "strings"
+)
+
+func isPalindrome(s string) bool {
+
+  s = strings.ToLower(s)
+
+	i, j := 0, len(s)-1
+	for i < j {
+		for i < j && !isChar(s[i]) {
+			i++
+		}
+		for i < j && !isChar(s[j]) {
+			j--
+		}
+		if s[i] != s[j] {
+			return false
+		}
+		i++
+		j--
+	}
+
+	return true
+}
+
+// 判断 c 是否是字符或者数字
+func isChar(c byte) bool {
+	if ('a' <= c && c <= 'z') || ('0' <= c && c <= '9') {
+		return true
+	}
+	return false
+}

Algorithms/125.Valid-Palindrome/Valid Palindrome_test.go

@@ -0,0 +1,41 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem125(t *testing.T) {
+
+	tcs := []struct {
+		s   string
+		ans bool
+	}{
+
+		{
+			"0p",
+			false,
+		},
+
+		{
+			"0",
+			true,
+		},
+
+		{
+			"race a car",
+			false,
+		},
+
+		{
+			"A man, a plan, a canal: Panama",
+			true,
+		},
+	}
+	fmt.Printf("------------------------Leetcode Problem 125------------------------\n")
+
+	for _, tc := range tcs {
+		fmt.Printf("【input】:%v       【output】:%v\n", tc, isPalindrome(tc.s))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/41.First-Missing-Positive/README.md

@@ -27,4 +27,11 @@ Output: 1
 
 Note:  
 
-Your algorithm should run in O(n) time and uses constant extra space.
+Your algorithm should run in O(n) time and uses constant extra space.
+
+## 题目大意
+
+找到缺失的第一个正整数。
+
+
+为了减少时间复杂度，可以把 input 数组都装到 map 中，然后 i 循环从 1 开始，依次比对 map 中是否存在 i，只要不存在 i 就立即返回结果，即所求。

Algorithms/412.Fizz-Buzz/README.md

@@ -30,3 +30,7 @@ Return:
     "FizzBuzz"
 ]
 ```
+
+## 题目大意
+
+3的倍数输出 "Fizz"，5的倍数输出 "Buzz"，15的倍数输出 "FizzBuzz"，其他时候都输出原本的数字。

Algorithms/88.Merge-Sorted-Array/README.md

@@ -5,4 +5,10 @@
 Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
 
 Note:
-You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
+You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
+
+## 题目大意
+
+合并两个已经有序的数组，结果放在第一个数组中，第一个数组假设空间足够大。要求算法时间复杂度足够低。
+
+为了不大量移动元素，就要从2个数组长度之和的最后一个位置开始，依次选取两个数组中大的数，从第一个数组的尾巴开始往头放，只要循环一次以后，就生成了合并以后的数组了。