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. Please note that 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.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

用hash查找：

# dictionary           
def twoSum2(self, numbers, target):
    dic = {}
    for i, num in enumerate(numbers):
        if target-num in dic:
            return [dic[target-num]+1, i+1]
        dic[num] = i

用二个指针：

class Solution(object):
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """        
        i = 0
        j = len(numbers)-1
        while i<j:
            if numbers[i]+numbers[j]==target:
                return [i+1, j+1]
            elif numbers[i]+numbers[j]>target:
                j -= 1
            else:
                i += 1
        return []     

二分：

class Solution(object):
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """ 
        for i in xrange(len(numbers)):
          l, r = i+1, len(numbers)-1
          tmp = target - numbers[i]
          while l <= r:
            mid = l + (r-l)//2
            if numbers[mid] == tmp:
                return [i+1, mid+1]
            elif numbers[mid] < tmp:
                l = mid+1
            else:
                r = mid-1

 二分优化：当target numer在数组表示的范围里，如果没有找到该数字，则循环结束一定有numbers[l]>target>numbers[r]，当初阿里面试就是找里某个数最近的数字。。。

class Solution(object):
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """                 
        def bsearch(nums, target, l, r): 
            if target < nums[l]: return (0, l)
            if target > nums[r]: return (0, r)
            while l <= r:
                mid = (l+r)>>1
                if nums[mid] == target:
                    return (1, mid)
                elif nums[mid] > target:
                    r = mid - 1
                else:
                    l = mid + 1
            return (0, r)
            # 3 4 5 => 3.5 final l=1, r=0 在范围里但是没有查找到
            # 3 4 5 => 4.5 final l=2, r=1 在范围里但是没有查找到
            # 5=>4 final l=0, r=-1 不在范围里
            # 4=>5 final l=1, r=0 不在范围里
        
        l = 0
        r = len(numbers)-1
        while l < r:
            found, j = bsearch(numbers, target-numbers[l], l+1, r)
            if found:
                return [l+1, j+1]
            else:
                l += 1
                r = j
        return []