The first question of leetcode, this method can achieve O(n) complexity solution
The question requires an int[], such as nums = [2, 7, 11, 15], and a target = 9.
If there are two numbers whose sum is the target value, for example nums[0] nums[1] = 2 7 = 9
return [0, 1].
When using the following solution, I have a little doubt, that is, a new hashmap is created, but no value is assigned to it. In this case, how do you achieve the question requirements?
public int[] twoSum(int[] numbers, int target) { int[] result = new int[2]; Map map = new HashMap(); for (int i = 0; i < numbers.length; i++) { if (map.containsKey(target - numbers[i])) { result[1] = i + 1; result[0] = map.get(target - numbers[i]); return result; } map.put(numbers[i], i + 1); } return result; } 
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Isn’t map.put() in the for loop an assignment? ? ?
The question requires that the sum of two numbers be the target value and a given value, then you must traverse at least two numbers. The realization of this situation is:
(1) If you initialize first, it will take algorithm time O(n); then traverse to find the first When the value is correct, the algorithm time is O(k), 0
1) Initialize the map first,
2) Traverse the first number 2, target - 2 = 9 - 2 = 7
3) Judge that 7 is also in the map and return the correct result.
Note: Traverse to the first correct number
(2) If it is not initialized, it will traverse to the second correct number before stopping. The algorithm time is O(k)(1
No
Key的情况下，HashMap.containsKey(Key)返回的是false不包括Key.There will be no null pointer error as you think.