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  • [leetcode]1. Two Sum两数之和

    Given an array of integers, return indices  of the two numbers such that they add up to a specific target.

    You may assume that each input would have exactly one solution, and you may not use the same element twice.

    Example:

    Given nums = [2, 7, 11, 15], target = 9,
    
    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].

    Solution1:  Brute-force

    Lock pointer i,  move pointer j to check if nums[j] == target - nums[i].

             

           

    code:

     1 /* Time Complexity: O(n*n)
     2    Space Complexity: O(1)
     3 */
     4 class Solution {
     5     public int[] twoSum(int[] nums, int target) {
     6         for (int i = 0; i < nums.length; i++) {
     7             for (int j = i + 1; j < nums.length; j++) {
     8                 if (nums[j] == target - nums[i]) {
     9                     return new int[] { i, j };
    10                 }
    11             }
    12         }
    13         return null;
    14     }
    15 }

    Solution2: HashMap

    Use a map to record the index of each element and check if target - nums[i] is in the map. if so, we find a pair whose sum is equal to target.

    code:

     1 /* Time Complexity: O(n)
     2    Space Complexity: O(n)
     3 */
     4 class Solution {
     5     public int[] twoSum(int[] nums, int target) {
     6         Map<Integer, Integer> map = new HashMap<>();
     7         for(int i = 0; i< nums.length; i++){
     8             if(map.containsKey(target - nums[i])){
     9                 return new int[]{i, map.get(target - nums[i])};
    10             }else{
    11                 map.put(nums[i], i);
    12             }
    13         }
    14         return null;
    15     }
    16 }

                          

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  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9080911.html
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