Two Sum

ゝ一世哀愁。 2022-05-30 12:06 164阅读 0赞

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].

此题可以用暴力求解，两个循环判断是否数组中存在两个数相加为目标值（不可自身与自身相加）

public int[] twoSum(int nums[], int target) {
    		 int k = 0;
    		 for (int i=0; i<nums.length; i++) {
    			 for (int j=i+1; j<nums.length; j++) {
    				 if (nums[j] == (target - nums[i])) {
    					 return new int[] {i, j};
    				 }
    			 }
    		 }
    		return null;
    	 }

我们可以看到这种求法时间复杂度为O(n^2) 空间复杂度为O(1)

现在我们打算用空间换时间，改用hashmap来求解

public int[] twoSum(int nums[], int target) {
    		HashMap<Integer, Integer> hm = new HashMap<>();
    		for (int i=0; i<nums.length; i++) {
    			hm.put(nums[i], i);
    		}
    		
    		for (int i=0; i<nums.length; i++) {
    			// cause the question needs us use different element
    			if (hm.containsKey(target - nums[i]) && (i != hm.get(target - nums[i]))) {
    				return new int[] {i, hm.get(target - nums[i]).intValue()};
    			}
    		}
    		return null;
    	}

这样时间和空间复杂度都成了O(n)