【leetcode hot 100 34】在排序数组中查找元素的第一个和最后一个位置

解法一：先用二分查找找到最左边的target，再顺序寻找最右边的target。

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n=nums.length;
        if(n==0){
            return new int[]{-1, -1};
        }
        int left=0, right=n-1, ret = n;
        while(left<=right){
            int mid = (left+right)/2; //取中间或者中间左边的那个数
            if(nums[mid]>=target){
                ret=mid; // 取到最左边的target
                right = mid-1;
            }
            else{
                left = mid+1;
            }
        }
        if(ret>=n){
            return new int[]{-1, -1};
        }
        if(nums[ret]==target){
            int i=ret+1;
            while(i<n){
                if(nums[i]!=target){
                    break;
                }
                i++;
            }
            return new int[]{ret, i-1};
        }
        return new int[]{-1, -1};
    }
}

注意：

• int mid = (left+right)/2 每次都取中间或者中间左边的那个数
• ret=mid 每次都取到最左边的target
• if(n==0){ return new int[]{-1, -1}; } 应对 nums=[] 的情况
• if(ret>=n){ return new int[]{-1, -1}; } 应对 target 应该在n位置上的情况

错误原因：时间复杂度应该为O(logn+n)

解法二：传入参数bool，确定每次取最左边的target还是取最右边的target

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int leftIdx = binarySearch(nums, target, true);
        int rightIdx = binarySearch(nums, target, false) - 1;
        if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] == target && nums[rightIdx] == target) {
            return new int[]{leftIdx, rightIdx};
        } 
        return new int[]{-1, -1};
    }

    public int binarySearch(int[] nums, int target, boolean lower) {
        int left = 0, right = nums.length - 1, ans = nums.length;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] > target || (lower && nums[mid] >= target)) {
                right = mid - 1;
                ans = mid;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
}