Question

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.

Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]

My Answer

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Best answer

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HashMap

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  1. /**
  2. * @param {number[]} nums
  3. * @return {boolean}
  4. */
  5. var containsDuplicate = function(nums) {
  6.   var map=[]
  7.   for(var i=0;i<nums.length;i++){
  8.     if(map[nums[i]]!==undefined){
  9.       return true
  10.     }
  11.     else{
  12.       map[nums[i]]=0
  13.     }
  14.   }
  15.   return false
  16. };

169. 多数元素

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  1. /**
  2.  * @param {number[]} nums
  3.  * @return {number}
  4.  */
  5. var majorityElement = function(nums) {
  6. const len=nums.length
  7. const map=new Array()
  8. if(len==1){
  9.     return nums[0]
  10. }
  11. for(let i=0;i<len;i++){
  12. if(map[nums[i]]==undefined){
  13.     map[nums[i]]=1
  14. }
  15. else{
  16.     map[nums[i]]++
  17.     if(map[nums[i]]>Math.floor(len/2))
  18.      return nums[i]
  19. }
  20. }
  22. };

26. 删除有序数组中的重复项

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  1. var removeDuplicates = function(nums) {
  2. const len=nums.length
  3. let slow=1;
  4. let fast=1;
  5. if (len === 0) {
  6.         return 0;
  7.     }
  8. while(fast<len){
  9.     if(nums[fast] !== nums[fast - 1]){
  10.       nums[slow] = nums[fast];
  11.     ++slow;  
  12.         }
  13.    ++fast
  14. }
  15. return slow;
  16. };