1. Two Sum（两个数求和）（leetcode）

分析：
1、第一次写是用双for循环写的，复杂度太高，pass
2、看答案，引用哈希表定位两者之间的对应关系，快了一倍。
运行时间：3ms            时间复杂度：o(n)

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

之前没看懂result赋值部分，要注意是将map中的值赋给result而不是nums中的。
3、考虑用双指针。这里涉及到给出的数组是否已经排好序，如果已经排好了，用双指针并行的时间复杂度是o(n/2);

题目167. Two Sum II - Input array is sorted（数组已排序的两个数求和）

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
方法一：双指针法
双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
时间复杂度：o（n/2)                运行时间：0ms           占用内存：35.8 MB

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int i=0;
        int j=numbers.length-1;
        while (i<j){
            int sum=numbers[i]+numbers[j];
            if(target==sum){
                return new int[]{i+1,j+1};
            }else if(target>sum){
                i++;
            }else {
                j--;
            }
        }
        return null;
    }
}

方法二：hashmap和上题同一个解法；

时间复杂度：o（n）              运行时间：2ms               占用空间：36mb

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-1;i++){
            if(map.containsKey(target-numbers[i])){
                result[0]=map.get(target-numbers[i])+1;
                result[1]=i+1;
            }
            map.put(numbers[i],i);
        }
        return result;
    }
}