34. Find First and Last Position of Element in Sorted Array

最新推荐文章于 2020-08-08 23:55:38 发布

原创最新推荐文章于 2020-08-08 23:55:38 发布 · 140 阅读

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博客围绕在升序整数数组中查找给定值的起始和结束位置展开，要求算法时间复杂度为O(log n)，若未找到目标值则返回特定结果，并给出了Java代码实现该查找功能。

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Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Java

class Solution {
public int[] searchRange(int[] nums, int target) {
int[] myArray = new int[]{-1,-1};
int start = -1;
int end = -1;

int first = 0;
int last =nums.length-1;
int middle =0 ;

if(nums.length==0){
return myArray;
}

while (first+1 < last){
middle = first +(last-first)/2;
if(nums[middle]==target){
last = middle;
}
else if(nums[middle]>target){
last = middle;
}
else if(nums[middle]<target){
first = middle;
}

}
if(nums[first]==target){
start=first;
}
else if(nums[last]==target){
start=last;
}
else{
return myArray;
}

myArray[0]=start;

first = 0;
last =nums.length-1;
middle =0 ;

while (first+1 <last){
middle = first +(last-first)/2;
if(nums[middle]==target){
first = middle;
}
else if(nums[middle]>target){
last = middle;
}
else if(nums[middle]<target){
first = middle;
}

}

if(nums[last]==target){
end=last;
}
else if(nums[first]==target){
end=first;
}
else{
return myArray;
}


myArray[1]=end;