daily 2019-06-20;594. Longest Harmonious Subsequence

daily/2019-06-20.md

@@ -0,0 +1,102 @@
+# 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
+};
+```

daily/README.md

@@ -71,6 +71,13 @@ 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`