描述
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have *exactly* one solution, and you may not use the same element twice.
You can return the answer in any order.
解题思路
使用HashMap存储nums[i]与target - nums[i]的关系映射，遍历数组元素，查找是否存在target - nums[i]索引，如果存在则返回。
代码实现
var twoSum = function(nums, target) {
let map = new Map();
for (let i = 0; nums.length; i++) {
if (map.has(nums[i])) {
return [map.get(nums[i]), i];
}
map.set(target - nums[i], i);
}
};描述
Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.
解题思路
用HashSet结构存储数组元素，判断集合的size值是否小于数组的长度。
代码实现
var containsDuplicate = function(nums) {
let set = new Set(nums);
return set.size < nums.length;
};描述
We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1.
Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.
A subsequence of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.
解题思路
和谐序列中最大数和最小数之差正好为1，使用HashMap结构存储元素出现次数，然后遍历其键值判断是否存在比其大1的键值，如果存在则相加两键值出现的次数，取所有和中的最大值。
代码实现
var findLHS = function(nums) {
let map = new Map();
for (let num of nums) {
map.set(num, (map.get(num) || 0) + 1);
}
let max = 0;
for (let num of map.keys()) {
if (map.has(num + 1)) {
max = Math.max(max, map.get(num) + map.get(num + 1));
}
}
return max;
};描述
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.
解题思路
使用HashSet数据结构保存数组元素，然后遍历其所有元素，内部循环遍历是否存在比单前值大1的元素，如果存在则连续序列长度+1，最后取所有可能连续序列中的最大长度。
代码实现
var longestConsecutive = function(nums) {
let set = new Set(nums);
let max = 0;
for (let num of set) {
if (set.has(num - 1)) continue;
let cur = num, cnt = 1;
while (set.has(cur + 1)) {
cur += 1;
cnt += 1;
}
max = Math.max(max, cnt);
}
return max;
};