167. Two Sum II

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.

My idea:先找出这个target的大致位置，用迭代比大小确定，然后再用迭代逐个相加

class Solution:
    def twoSum(self, numbers,target):
        i = 0
        while (numbers[i] < target):
            i = i + 1
            if(max(numbers)<=target):
                i=len(numbers)-1
                break
        if(i==0):
            return [1,2]
        b = 1
        flag = 0
        c = []
        for a in range(i):
            while (flag == 0 ):

                if (numbers[a] + numbers[b] == target):
                    flag = 1
                    c.append(a + 1)
                    c.append(b + 1)
                    return c
                b = b + 1
                if(b >i):
                    break
            b=a+2
        return c

你以为这样就结束了？

不，这样耗时太长，所以失败了。。。因为碰到了一个2500char的测试用例

而且这样写要考虑的情况太多，不简便，不推荐。

其实学的就是双指针，所以直接双指针就完事了。。

class Solution:
    def twoSum(self, numbers,target):
        a=0
        b=len(numbers)-1
        while(a<b):
            if(numbers[a]+numbers[b]==target):
                c=[a+1,b+1]
                return c
            elif(numbers[a]+numbers[b]<target):
                a=a+1
            else:
                b=b-1

执行用时 : 60 ms, 在Two Sum II - Input array is sorted的Python3提交中击败了62.54% 的用户

内存消耗 : 13.4 MB, 在Two Sum II - Input array is sorted的Python3提交中击败了95.42% 的用户