[LeetCode 001] Two Sum

Two Sum

• 在HashMap中存储numbers[i]尚需要都少数值达到target，以及下表i

Implementation

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
    	HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    	int[] result = new int[2];
    	for (int i = 0; i < numbers.length; i++) {
    		if (map.containsKey(numbers[i])) {
    			int index = map.get(numbers[i]);
    			result[0] = index;
    			result[1] = i;
    			break;
    		} else {
    			map.put(target - numbers[i], i);
    		}
    	}
    	return result;
    }
}

Two Sum II - Input array is sorted

• 设置两个指针j指向最大值，i指向最小值
  • sum > target: jdecrease
  • sum < target: iincrease

Implementation

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int left = 0;
        int right = numbers.length - 1;
        int[] result = new int[2];
        while (left < right) {
            int sum = numbers[left] + numbers[right];
            if (sum == target) {
                result[0] = left + 1;
                result[1] = right + 1;
                break;
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
        return result;
    }
}

Two Sum III - Data structure design

• 设置一个ArrayList保存出现过的不重复的数值。
  • 遍历的时候比使用HashMap的keySet()更快
• 设置一个HashMap保存出现过得不重复的数值，以及出现次数。
• find()函数
  • 当value需要两个相同的数值得到时，判断该数值出现的次数是否大于1
  • 否则，判断HashMap是否包含这两个数值

Implementation

public class TwoSum {
    private HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    private ArrayList<Integer> list = new ArrayList<Integer>();
    // Add the number to an internal data structure.
	public void add(int number) {
	    if (map.containsKey(number)) {
	        map.put(number, map.get(number) + 1);
	    }
	    else {
	        map.put(number, 1);
	        list.add(number);
	    }
	}

    // Find if there exists any pair of numbers which sum is equal to the value.
	public boolean find(int value) {
	    for (int i = 0; i < list.size(); i++) {
	        int num1 = list.get(i);
	        int num2 = value - num1;
	        if ((num1 == num2 && map.get(num1) > 1) || (num1 != num2 && map.containsKey(num2))) {
	            return true;
	        }
	    }
	    return false;
	}
}