- 题目描述:
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
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2. 解题思路:
因为这题的数组是一个有序的数组,首先想到的应该是采用二分查找的方式进行查找,然后再分别找到出现的初试位置和最后出现的位置。
3. 解题代码:
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int length = nums.size();
//如果target < nums[0] 或者target > nums[nums.size()-1] 表明不存在
if(length == 0 || target < nums[0] || target > nums[length-1] ){
return {
-1,-1};
}
int left =0,right = length-1;
int mid = 0;
while(left <= right){
mid = (left + right)/2;
if(nums[mid] == target){
break;
}
if(nums[mid] < target){
while(mid + 1 < length - 1 && nums[mid] == nums[mid + 1]){
mid ++;
}
left = mid + 1 ;
}
if(nums[mid] > target){
while(mid-1 >= 1 && nums[mid] == nums[mid-1] ){
mid--;
}
right = mid -1 ;
}
}
if(left > right){
return {
-1,-1};
}
int temp = mid;
while(mid + 1 < length && nums[mid] == nums[mid + 1] ){
mid ++;
}
while(temp-1 >= 0 && nums[temp] == nums[temp-1] ){
temp--;
}
return {
temp,mid};
}
};
- tips:
一般对有序的数据结构进行查找一般都是采用二分查找的方式来进行查找。