【LeetCode OJ 34】Search for a Range
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题目链接:https://leetcode.com/problems/search-for-a-range/
题目:Given a sorted array of integers, 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]
.
解题思路:题意为在一个有序数组中寻找指定值的起始位置和结束位置,因为题目要求时间复杂度为O(log n),故而想到使用二分查找法。
示例代码:
public class Solution { public int[] searchRange(int[] nums, int target) { int[] result=new int[]{-1,-1}; int start=0; int end=nums.length-1; while(start<=end) { int middle=(start+end)>>1; int middleValue=nums[middle]; if(middleValue==target) { //找到目标值后,先搜索其左边有没有与目标值相等的数,取其最左边的索引值放入result中,同理再搜索其最右边 int i=middle; while((i-1)>=0&&nums[i-1]==target) { i--; } result[0]=i; int j=middle; while((j+1)<nums.length&&nums[j+1]==target) { j++; } result[1]=j; return result; } else if(target<middleValue) { //小于中值时在中值前面找 end=middle-1; } else { //大于中值在中值后面找 start=middle+1; } } return result; }}
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