Two Sum
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].
思路1
双层循环遍历数组，外层循环取
first数字，内层循环取second数字，它们的和等于target则返回.-
实现代码
private int[] twoSum(int[] arrays, int target) { int length = arrays.length; for (int first = 0; first < length; first++) { for (int second = first + 1; second < length; second++) { if (target == arrays[first] + arrays[second]) { return new int[]{first, second}; } } } return null; }
思路2
用
HashMap存储key为数组中每个数字，value为该数字出现的位置。
单层循环，每次取一个数字arrays[index]，然后判断HashMap中是否存在key = target - arrays[index]，如果存在则返回，不存在则将取到的数字放入HashMap中.-
实现代码
private int[] twoSum(int[] arrays, int target) { HashMap<Integer, Integer> map = new HashMap<>(length); for (int index = 0; index < arrays.length; index++) { if (map.containsKey(target - arrays[index])) { return new int[]{map.get(target - arrays[index]), index}; } map.put(arrays[index], index); } return null; }
