1.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].

public class Solution {    public int[] twoSum(int[] nums, int target) {        int[] solution=new int[2];        for(int i=0;i<nums.length-1;i++){            for(int j=i+1;j<nums.length;j++){                if(nums[i]+nums[j]==target){                    solution[0]=i;                    solution[1]=j;                    return solution;                }            }        }        return solution;    }    }

上面这种方法是遍历，复杂度比较高，嗯，编程能力比较小白

其他方法我看到有一种复杂度为O(n)的，采用了HashMap；containsKey（value）表示查找map中含有括号中的value；get（value）表示获得map中相应value的key值；put（value,key）把值放入map中

public class Solution{    public int[] twoSum(int[] numbers, int target) {        int[] result = new int[2];        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        for (int i = 0; i < numbers.length; i++) {            if (map.containsKey(target - numbers[i])) {                result[1] = i;                result[0] = map.get(target - numbers[i]);                return result;            }            map.put(numbers[i], i);        }        return result;    }}

相关链接：https://discuss.leetcode.com/topic/2447/accepted-java-o-n-solution

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