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]
java代码：
1 public class Solution { 2 public int[] twoSum(int[] nums, int target) { 3 for(int i=0;i<nums.length;i++){ 4 for(int j=1;j<nums.length;j++){ 5 if(target==nums[i]+nums[j]){ 6 return new int[]{i,j}; 7 } 8 } 9 } 10 throw new RuntimeException("No such this indices"); 11 } 12 }
1 public class Solution { 2 public int[] twoSum(int[] numbers, int target) { 3 int[] result = new int[2]; 4 Map<Integer, Integer> map = new HashMap<Integer, Integer>(); 5 for (int i = 0; i < numbers.length; i++) { 6 if (map.containsKey(target - numbers[i])) { 7 result[1] = i; 8 result[0] = map.get(target - numbers[i]); 9 return result; 10 } 11 map.put(numbers[i], i + 1); 12 } 13 return result; 14 } 15 }
1 public int[] twoSum(int[] nums, int target) { 2 Map<Integer, Integer> map = new HashMap<>(); 3 for (int i = 0; i < nums.length; i++) { 4 map.put(nums[i], i); 5 } 6 for (int i = 0; i < nums.length; i++) { 7 int complement = target - nums[i]; 8 if (map.containsKey(complement) && map.get(complement) != i) { 9 return new int[] { i, map.get(complement) }; 10 } 11 } 12 throw new IllegalArgumentException("No two sum solution"); 13 }
原文：http://www.cnblogs.com/lwy19998273333/p/6404541.html