toSum

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 }

 

posted @ 2017-02-16 10:30  霓羽决奕  阅读(449)  评论(0编辑  收藏  举报