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].
兩個(gè)循環(huán)遍歷，取兩個(gè)序號(hào)
public class Solution { public int[] twoSum(int[] nums, int target) { int[] result = new int[2]; for(int i = 0; i < nums.length; i++) { for (int j = i + 1; j < nums.length; j++) { if (nums[i] + nums[j] == target) { result[0] = i; result[1] = j; return result; } } } return result; }}Runtime: 44 ms
利用hashmap，key存放數(shù)值，value存放出現(xiàn)的位置。從前到后進(jìn)行遍歷，將target值減去當(dāng)前的值，看是否存在map中，
若存在map中則取出相應(yīng)的標(biāo)號(hào)，退出。
public class Solution { public int[] twoSum(int[] nums, int target) { int[] result = new int[2]; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for(int i = 0; i < nums.length; i++) { int num = target - nums[i]; if (map.containsKey(num)) { result[0] = map.get(num); result[1] = i; return result; } map.put(nums[i], i); } return result; }}Runtime: 12 ms
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