<easy>LeetCode Problem -- 167. Two Sum II - Input array is sorted

描述：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

分析：给定了一个数组和一个target，要求输出数组中和为target的值的下标。

思路一：循环遍历，如果存在一个值和target-nums[i]相等，则将这两个值的下标输出。不出所料，果然超时。

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

思路二：由于思路一会产生超时，因此需要一种时间复杂度为O(n)的算法。仔细分析题目发现题目中强调了两点：

1.数组升序

2.和为target的值一定存在

基于这两点我们可以按照两个数和的大小关系来进行判断。从前后两个方向同时进行计算并比较和的大小。

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