Screenshot from 2016-03-01 21:28:37.png

My code:

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

不知道这道题目有什么意义。跟2sum不是差不多嘛？

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        if (numbers == null || numbers.length == 0) {
            return null;
        }
        
        int begin = 0;
        int end = numbers.length - 1;
        while (begin < end) {
            int sum = numbers[begin] + numbers[end];
            if (sum > target) {
                end--;
            }
            else if (sum < target) {
                begin++;
            }
            else {
                int[] ret = new int[2];
                ret[0] = begin + 1;
                ret[1] = end + 1;
                return ret;
            }
        }
        
        return null;
    }
}

reference:
https://leetcode.com/articles/two-sum-ii-input-array-sorted/

看到这道题目，我首先想到了两个解法。
第一个就是用hashtable
time: O(n)
space: O(n)

第二种是用 遍历 + binary search
time: O(n logn)
space: O(1)

有没有种算法，
time: O(n)
space: O(1)

当然是有的。这种排好序的数列，就是天生为了双指针而存在的。

Anyway, Good luck, Richardo! — 09/02/2016