二分查找-leetcode34

给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。

示例 1：

5,7,7,8,8,10]

示例 2：

5,7,7,8,8,10]

示例 3：

输入：nums = [], target = 0
输出：[-1,-1]

提示：

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums 是一个非递减数组
• -109 <= target <= 109

思路：二分法，找到数之后分别左右寻找边界

class Solution {
    public int[] searchRange(int[] nums, int target) {
       return new int[]{searchLeftBound(nums,target),searchRightBound(nums,target)};
    }

    int searchLeftBound(int[] nums, int target){
        int left =0;
        int right = nums.length-1;
        while(left<=right){
            int mid = left + (right-left)/2;
            if(nums[mid]<target){
                left = mid+1;
            }else if(nums[mid]>target){
                right=mid-1;
            }else if (nums[mid]==target){
                right = mid-1;
            }
            //System.out.println("left: "+left);
        }

        if(left>=nums.length || nums[left]!=target){
            return -1;
        }
        return left;

    }

    int searchRightBound(int[] nums, int target){
        int left =0;
        int right = nums.length-1;
        while(left<=right){
            int mid = left + (right-left)/2;
            if(nums[mid]<target){
                left = mid+1;
            }else if(nums[mid]>target){
                right=mid-1;
            }else if (nums[mid]==target){
                left = mid+1;
            }
        }

        if(right<0 || nums[right]!=target){
            return -1;
        }
        return right;

    }
}