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;}
}