在排序数组中查找元素的第一个和最后一个位置

解题：利用二分找到一个相等的元素，之后再左右边界

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n=nums.length;
        int l=0,r=n-1;
        int start=-1,end=-1;
        while(l<=r){
            int mid=(r+l)/2;
            if(nums[mid]==target){
                start=mid;
                end=mid;
                int temp=mid;
                while(nums[mid]==target){
                     mid--;
                     if(mid==-1) break;
                }
                start=mid+1;
                while(nums[temp]==target){
                    temp++;
                    if(temp==n) break;
                } 
                end=temp-1;
                break;
            }else if(nums[mid]>target){
                r=mid-1;
            }else{
                l=mid+1;
            }
        }
        return new int[]{start,end};

    }
}

在找到一个一个等于target数时做两种不同的选择，r=mid-1,l=mid+1可以找出左右边界

class Solution {
    public int[] searchRange(int[] nums, int target) {
        return new int[]{binarySearchLeftBound(nums, target),binarySearchRightBound(nums, target)};
        
    }
    public int binarySearchLeftBound(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = l + ((r - l) >> 1);
            if (nums[mid] < target)  {
                l = mid + 1;
            } else if (nums[mid] > target) {
                r = mid - 1;
            } else if (nums[mid] == target) {
                r = mid - 1;
            }
        }        
        if (l >= nums.length || nums[l] != target) return -1;
        return l;
    }
    public int binarySearchRightBound(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
        while (l <= r) {
            int mid = l + ((r - l) >> 1);
            if (nums[mid] < target)  {
                l = mid + 1;
            } else if (nums[mid] > target) {
                r = mid - 1;
            } else if (nums[mid] == target) {
                l = mid + 1;
            }
        }
        if (r < 0 || nums[r] != target) return -1;
        return r;
    }
}