在LeetCode上做的第一道题,记录一下:
Given an array of integers, 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
解题思路:
借鉴《编程之美》2.12 快速寻找满足条件的两个数:
对数组进行排序(注意题目要求,需要返回的是index,所以这里需对数组进行copy,在数组副本上进行排序),然后用双向索引线性搜索出一对解。思路大致如下:
升序排列后,设有序数组为a,目标求和数为t,索引i<索引j(即a[i]<a[j])则有如下规律:
若a[i] + a[j] < t, 则有a[i]+a[k]<t (k<j)
若a[i] + a[j] > t, 则有a[k]+a[j]>t (k>i)
也就是说,当a[i]和a[j]之和小于t时,我们只需要保持j不变,右移i进行下一次尝试。反之则保持i不变,左移j进行下一次尝试。这是一个线性扫描过程,复杂度为O(n)。
当找出这样一对解后,剩下的问题就是到原数组里面找出他们的索引即可。因为题目中已告知一组输入只会有唯一的解,所以这里不需要考虑多组解可能出现的重复问题,所以,我的思路是对原数组进行一次正向遍历,遇到一个元素与解中的数值相同即记录下此时的index,当记录了两个index后,退出遍历即可。
代码如下:
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int> temp = numbers;
sort(temp.begin(),temp.end());
int i=0,j = temp.size()-1;
for(;i<j;){
if(temp[i]+temp[j] > target) {
j--;
continue;
}else if((temp[i]+temp[j])<target){
i++;
continue;
}else break;
}
vector<int> answer;
int num1 = temp[i];
int num2 = temp[j];<span style="font-family: Arial, Helvetica, sans-serif;"> </span>
<pre name="code" class="cpp"><span style="white-space:pre"> </span>for(int iter=0;iter<numbers.size();iter++){
if(numbers[iter] == num1 || numbers[iter] == num2){
answer.push_back(iter+1);
}
}
<span style="font-family: Arial, Helvetica, sans-serif;"> </span>
return answer; }};
上述代码运行结果是58ms, 如果将最后寻找index的过程分成两个循环来写,可以在找到所有index后即退出循环,会更快一点,修改如下:
/* for(int iter=0;iter<numbers.size();iter++){
if(numbers[iter] == num1 || numbers[iter] == num2){
answer.push_back(iter+1);
}
}*/
//改成这样:
int iter =0,next;
for(;iter<numbers.size()-1;iter++){
if(numbers[iter] == num1){
next = num2;
break;
}
if(numbers[iter] == num2){
next = num1;
break;
}
}
answer.push_back(++iter);
for(;iter<numbers.size()-1;iter++){
if(numbers[iter] == next) break;
}
answer.push_back(iter+1);
这样的运行结果是48ms