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]

思路：題目要求時間復雜度為O(log n)，且數組有序，可用二分查找法。但數組中有重復的元素，因此需要對二分查找作一點修改。

代碼：

    //二分查找 -- 有重復元素  查找最開頭與最結尾的目標值
    public static int[] searchRange(int[] nums, int target) {
        
        int[] res = new int[2];
        List<Integer> l = new
 
 ArrayList<>();
        
        int left = 0;
        int right = nums.length-1;
        
        while(left<=right){
            
            int mid = (left+right)/2;
            if(nums[mid]==target){
                int start = mid;
                int end = mid;
                //獲取start的索引
 

                while(start>=0){
                    if(nums[start]!=target){
                        break;
                    }
                    start--;
                }
                l.add(start+1);
                //獲取end的索引
                while(end<=nums.length-1){
                    if(nums[end]!=target){
                        break;
                    }
                    end++;
                }
                l.add(end-1);
                
                break;
            }else if(nums[mid]<target){
                left = mid+1;
            }else{
                right = mid-1;
            }
        }
        if(l.size()==0){
            res[0] = -1;
            res[1] = -1;
        }else{
            res[0] = l.get(0);
            res[1] = l.get(1);
        }
        
        return res;
     }

當然，如果不考慮時間復雜度，也可用暴力法。

代碼：

    //暴力循環
     public static int[] searchRange(int[] nums, int target) {
        
        int[] res = new int[2];
        List l = new ArrayList<>();
         
        for (int i = 0; i < nums.length; i++) {
            if(nums[i] == target){
                l.add(i);
            }
        }
        
        if(l.size()==0){
           res[0] = res[1] =-1;
        }else if(l.size()==1){
            res[0] = res[1] = (int) l.get(0);
        }else{
            res[0] = (int) l.get(0);
            res[1] = (int) l.get(l.size()-1);
        }
        
         return res;
     }

Find First and Last Position of Element in Sorted Array（數組中查找第一個元素和最後一個元素）