[LeetCode] 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.

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从numbers中找两数之和为target，返回下标

这道题我是通过Binary Search的标签进入的，所以想都没想直接写了，最后虽然AC了但成绩并不好，都是菜鸟的借口

int binarySearch(vector<int>& vec, int low, int high, int target)
{
    while (low <= high)
    {
        int mid = (low + high) / 2;

        if (vec[mid] == target)
        {
            return mid;
        }
        else if (vec[mid] > target)
        {
            high = mid-1;
        }
        else if (vec[mid] < target)
        {
            low = mid + 1;
        }
    }

    return -1;
}

vector<int> twoSum(vector<int>& numbers, int target) {
    for (int i = 0; i < numbers.size()-1; i++)
    {
        int rest = target - numbers[i];

        int index = binarySearch(numbers, i+1, numbers.size(), rest);
        if (index != -1)
        {
            return vector<int> {i+1, index+1};
        }
    }
}

题目说了，vector是排好序的，所以可以更快找到答案

用low和high指向numbers的两端，low++使两数之和增加，high--使两束之和减少，逐渐逼近target

vector<int> twoSum(vector<int>& numbers, int target) {
    vector<int> result;
    int low = 0;
    int high = numbers.size() - 1;

    while (low <= high)
    {
        int sum = numbers[low] + numbers[high];
        if (sum == target)
        {
            return vector<int> {low + 1, high + 1};
        }
        else if (sum > target)
        {
            high--;
        }
        else if (sum < target)
        {
            low++;
        }
    }
}