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]
题目
有序数组中找重复数字出现和结束的位置
思路
和33题很像，一样是二分法，这一次稍微简单点，如果nums[mid]=target，则两个独立的while一个往回找，一个往后找，找到边界，存储索引即可。
代码
public int[] searchRange(int[] nums, int target) {
int len = nums.length;
int l=0;
int r=len-1;
int ind1,ind2;
int[] res = new int[2];
while (l<=r) {
int mid=(l+r)/2;
if (nums[mid]==target) {
ind1=mid;
ind2=mid;
while (ind1>=0&&nums[ind1]==target) {
ind1--;
}
while (ind2<=len-1&&nums[ind2]==target) {
ind2++;
}
res[0]=ind1+1;
res[1]=ind2-1;
return res;
}else if (nums[mid]>target) {//证明得去左边找
r=mid-1;
}else {
l=mid+1;
}
}
if (l>r) {
res[0]=-1;
res[1]=-1;
}
return res;
}
