LeetCode刷题EASY篇Two Sum II - Input array is sorted

题目

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.

我的思路

首先遍历，存入map，第二次遍历，判断target-current value 元素是否在map中，在，返回索引。

之前有two sum的题目，数组是无序的，我原来也是采用这个思路，但是可以优化，不需要遍历两次，一次就够，看下面的代码

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}

这个思路肯定可行，但是无法利用有序这个特点，应该不是最优解法。

利用两个指针遍历，根据当前sum决定指针的移动情况，测试通过，但是只是超过了1%

的提交，时间复杂度太高。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        int p = 0;
        int q = 1;
         while (p < nums.length || q < nums.length) {
            if (q > nums.length - 1) {
                p++;
                q = p + 1;
            }
            int sum = nums[p] + nums[q];
            if (sum < target) {
                q++;
            }
            if (sum == target) {
                res[0] = p + 1;
                res[1] = q + 1;
                return res;
            }
            if (sum > target) {
                p++;
                q = p + 1;
            }


        }
        return res;
    }
}

优化解法

两个指针没有错，应该从两头开始。这样效率就会高很多，看了其他人的思路，修改一下：

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        int p = 0;
        int q = nums.length-1;
         while (p < q) {
            int sum = nums[p] + nums[q];
            if (sum < target) {
                p++;
            }
            if (sum == target) {
                res[0] = p + 1;
                res[1] = q + 1;
                return res;
            }
            if (sum > target) {
                q--;
            }


        }
        return res;
    }
}