[LeetCode] 167. Two Sum II - Input array is sorted 两数和 II - 输入是有序的数组

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.

思路：

与1. Tow Sum类似，这题的输入是有序数组，限定了一定会有解，用双指针来做，定义左右两个指针，左指针指向第一个数，右指针指向最后一个数，然后用这两个数的和与Target比较，如果比Target小，左指针向右移一位，如果比Target大，右指针向左移一位。然后再进行比较，直到找到或者两个指针相遇为止。

Time: O(n)  Space: O(1)

Java:

public class Solution {
public int[] twoSum(int[] numbers, int target) {
if(numbers==null || numbers.length < 1) return null;
int i=0, j=numbers.length-1;

while(i<j) {
int x = numbers[i] + numbers[j];
if(x<target) {
++i;
} else if(x>target) {
--j;
} else {
return new int[]{i+1, j+1};
}
}
return null;
}
}

Python：

class Solution:
def twoSum(self, nums, target):
start, end = 0, len(nums) - 1

while start != end:
sum = nums[start] + nums[end]
if sum > target:
end -= 1
elif sum < target:
start += 1
else:
return [start + 1, end + 1]

C++：

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int l = 0, r = numbers.size() - 1;
while (l < r) {
int sum = numbers[l] + numbers[r];
if (sum == target) return {l + 1, r + 1};
else if (sum < target) ++l;
else --r;
}
return {};
}
};

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