leetcode题目例题解析(三)
Search for a Range
题目描述:
Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm’s runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10]
and target value 8
return [3, 4].
题意解析:
这道题的主要意思就是一个简单的查找,而且目标数组就已经排好序的,所以最快的办法就是用二分查找,要求中是O(log n)的复杂度,一看就知道使用二分查找,这里有一些要注意的,就是如果没有找到,就要返回-1,-1。其他地方没有什么难点和坑点。
解题思路:
直接二分查找,用两个迭代器分别代表查找的开始和结尾,保持开始>=结尾就行了
代码:
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int start, end, mid;
int len = nums.size();
int left = 0;
int right = len - 1;
while (left <= right) {
mid = (left + right)/2;
if (nums[mid] == target) {
start = mid;
end = mid;
//这里注意不要越界,不然可能会导致错误的结果
while(start >= 0&&nums[start] == target) start--;
while(end < len&&nums[end] == target) end++;
start++;
end--;
break;
}
if (nums[mid] > target) {
right = mid - 1;
}
if (nums[mid] < target) {
left = mid + 1;
}
}
if (left > right)
start = end = -1;
vector<int> res;
res.push_back(start);
res.push_back(end);
return res;
}
};
原题目链接:
https://leetcode.com/problems/search-for-a-range/description/