合并master

Author: luzhipeng

Diff for: daily/2019-06-17.md

@@ -0,0 +1,84 @@
+# 毎日一题 -  744. find smallest letter greater than target
+
+## 信息卡片
+* 时间：2019-06-17
+* 题目链接：https://leetcode.com/problems/find-smallest-letter-greater-than-target/
+* tag：`Array`
+
+## 题目描述
+```
+Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.
+
+Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.
+
+Examples:
+    Input:
+    letters = ["c", "f", "j"]
+    target = "a"
+    Output: "c"
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "c"
+    Output: "f"
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "d"
+    Output: "f"
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "g"
+    Output: "j"
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "j"
+    Output: "c"
+
+    Input:
+    letters = ["c", "f", "j"]
+    target = "k"
+    Output: "c"
+Note:
+    letters has a length in range [2, 10000].
+    letters consists of lowercase letters, and contains at least 2 unique letters.
+    target is a lowercase letter.
+```
+
+## 思路
+二分查找，提高速度
+要求是查找某一个元素，又是在有序的集合中。
+所以我们可以用二分查找
+1. 排除两种情况；target 小于首元素|| target 大于等于尾元素 => 目标都是首元素
+2. 当target>=letters[mid] 时(我们要的值一定在右边)，调整左区间 min = mid+1;
+3. 当target< letters[mid] 时，调整右区间 max = mid-1;
+4. 循环终止条件是 min > max; 最终返回min位置元素
+
+## 建议
+在leetcode上找一个数组稍微长一点的测试用例，在纸上画出整个过程；对理解很有帮助
+
+## 参考答案
+```js
+/**
+ * @param {character[]} letters
+ * @param {character} target
+ * @return {character}
+ */
+var nextGreatestLetter = function(letters, target) {
+    const length = letters.length
+    let min = 0;
+    let max = length - 1;
+    if(target >= letters[length-1] || target < letters[0]) return letters[0];
+    while(min <= max) {
+        const mid = (max+min) >> 1
+        if(target >= letters[mid]) {
+            min = mid + 1;
+        } else {
+            max = mid - 1;
+        }
+    }
+    return letters[min]
+};
+```

Diff for: daily/2019-06-20.md

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+# 毎日一题 -  594. Longest Harmonious Subsequence
+
+## 信息卡片
+* 时间：2019-06-20
+* 题目链接：https://leetcode.com/problems/longest-harmonious-subsequence/
+* tag：`Array`
+
+## 题目描述
+```
+We define a harmounious array as an array where the difference between its maximum value and its minimum value is exactly 1.
+
+Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.
+
+Example 1:
+
+  Input: [1,3,2,2,5,2,3,7]
+  Output: 5
+  Explanation: The longest harmonious subsequence is [3,2,2,2,3].
+```
+
+## 思路
+1. 将数组中的值作为一个对象中的属性，出现的次数就是属性值
+2. 属性差一的值相加，获取最大的，否则返回0；
+
+## 参考答案
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ * 使用ES6中的Map
+ */
+var findLHS = function(nums) {
+    if(!nums.length) return 0;
+    const map = new Map();
+    let max = 0;
+    for(let i = 0; i<nums.length; i++) {
+      let target = nums[i]
+      if (map.has(target)) {
+        map.set(target, map.get(target)+1);
+      } else {
+        map.set(target, 1);
+      }
+    }
+    for (let key of map.keys()) {
+      if(map.has(key+1)) {
+        max = Math.max(map.get(key)+map.get(key+1), max);
+      }
+    }
+    return max
+};
+```
+
+### 其它优秀解法
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ * for...in遍历
+ */
+var findLHS = function(nums) {
+    if(!nums.length) return 0;
+    const counts = {};
+    let max = 0;
+    for(let i = 0; i<nums.length; i++) {
+      if (counts[nums[i]]) {
+        counts[nums[i]] += 1;
+      } else {
+        counts[nums[i]] = 1;
+      }
+    }
+    for (let key in counts) { // for...in性能低
+      if(counts[+key+1]) {
+        max = Math.max(counts[key]+counts[+key+1], max)
+      }
+    }
+    return max
+};
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ * 普通遍历
+ */
+var findLHS = function(nums) {
+    if(!nums.length) return 0;
+    const counts = {};
+    let max = 0;
+    for(let i = 0; i<nums.length; i++) {
+      if (counts[nums[i]]) {
+        counts[nums[i]] += 1;
+      } else {
+        counts[nums[i]] = 1;
+      }
+    }
+    for (let i = 0; i < nums.length; i++) { // 有多余的无效遍历
+      if (counts[nums[i] + 1]) {
+        max = Math.max(max, counts[nums[i]] + counts[nums[i] + 1]);
+      }
+    }
+    return max
+};
+```

Diff for: daily/2019-07-25.md

@@ -0,0 +1,92 @@
+# 毎日一题 - 9. Palindrome number
+
+## 信息卡片
+
+- 时间：2019-07-25
+- 题目链接：https://leetcode.com/problems/palindrome-number/submissions/
+- tag：`Math`
+
+## 题目描述
+
+```
+Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
+
+Example 1:
+
+  Input: 121
+  Output: true
+
+Example 2:
+
+  Input: -121
+  Output: false
+  Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
+
+Example 3:
+
+  Input: 10
+  Output: false
+  Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
+
+Follow up:
+
+  Coud you solve it without converting the integer to a string?
+```
+
+## 参考答案
+
+转成字符串方式
+1. 负数都是非回文数，10的整数倍不是回文。
+2. 将数字转为字符串，再逆序排列字符串。两者比较，相等就是回文数。
+
+直接操作整数方式
+1. 复制x到temp;
+2. 取temp末尾数字，方式为temp与10的求余；组成新数reverse;
+3. 每取完一位,temp缩小10倍并且去掉小数。
+4. reverse要`先扩大十倍`再加上取下来的数
+5. 当temp === 0时，表示已经取完；reverse与x比较
+
+
+参考JavaScript代码：
+
+```js
+/**
+ * @param {number} x
+ * @return {boolean}
+ * 转成字符串
+ */
+var isPalindrome = function(x) {
+    if(x < 0 ) return false;
+    if(x === 0) return true;
+    if(x % 10 === 0) return false;
+    let reverse = '';
+    let str = String(x);
+    for(let i = str.length - 1; i >= 0; i--) {
+      reverse += str[i]
+    }
+    return str === reverse
+};
+
+/**
+ * @param {number} x
+ * @return {boolean}
+ * 不转成字符串
+ */
+var isPalindrome = function(x) {
+    if(x < 0 ) return false;
+    if(x === 0) return true;
+    if(x % 10 === 0) return false;
+    let temp = x;
+    let reverse = 0;
+    while(temp > 0) {
+      let num = temp % 10;
+      temp = (temp - num)/10; // 或 temp = (temp / 10) >> 0,去除小数位
+      reverse = reverse * 10 + num
+    }
+    return x === reverse
+};
+```
+
+## 优秀解答
+
+> 暂缺

Diff for: daily/README.md

@@ -60,6 +60,12 @@ tag: `tree`
 
 时间: 2019-06-14
 
+### [744.find-smallest-letter-greater-than-target](./2019-06-17.md)
+
+tag：`Array` `binary search`  
+
+时间：2019-06-17
+
 ### [letter-combinations-of-a-phone-number](./2019-06-18.md)
 
 tag: `backtrack`
@@ -72,6 +78,12 @@ tag: `sql`
 
 时间: 2019-06-19
 
+### [594.longest-harmonious-subsequence](./2019-06-20.md)
+
+tag：`Array`  
+
+时间：2019-06-20
+
 ### [nth-highest-salary](./2019-06-21.md)
 
 tag: `sql`
@@ -144,6 +156,12 @@ tag：`发散思维`
 
 时间: 2019-07-24
 
+### [9.palindrome-number](./2019-07-25.md)
+
+tag：`Math`
+
+时间: 2019-07-25
+
 ### [将帅问题](./2019-07-26.md)
 
 tag：`数据压缩`