[LeetCode167]Two Sum II - Input array is sorted

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.

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

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这题跟two sum 不是一样吗？当时做的时候就 sort original array first。

然后come up with a O(n) method:

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int lo = 0, hi = numbers.size()-1;
vector<int> res;
while(lo<=hi){
if(numbers[lo] + numbers[hi] < target) ++lo;
else if(numbers[lo] + numbers[hi] > target) --hi;
else {
res.push_back(lo+1);
res.push_back(hi+1);
break;
}
}
return res;
}
};

然后看看tag: binary search O(nlogn), less efficient.

但我不知道怎么写，懒：

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
if(numbers.empty()) return {};
for(int i=0; i<numbers.size()-1; i++) {
int start=i+1, end=numbers.size()-1, gap=target-numbers[i];
while(start <= end) {
int m = start+(end-start)/2;
if(numbers[m] == gap) return {i+1,m+1};
else if(numbers[m] > gap) end=m-1;
else start=m+1;
}
}
}
};