LeetCode34. 在排序数组中查找元素的第一个和最后一个位置

思路

方法一就是先找的target然后从其low开始找第一个target，从起hight位置倒着开始找第一个target。

方法二将目标数target改为target-0.5和target+0.5.分两次二分，每次返回其low位置。

这种思路很巧妙。 

代码

方法一：

class Solution {
    public int[] searchRange(int[] nums, int target) {
       
        int lo = 0;
        int hi = nums.length-1;
        int mid = 0;
        int[] res = new int[]{-1,-1}; 
        while(lo<=hi){
            mid = lo + (hi-lo)/2;
            if(nums[mid] == target){
                for(int i = lo;i<=hi;i++){
                    if(nums[i] == target){
                        res[0] = i;
                        break;
                    }
                }
                for(int j = hi;j>=lo;j--){
                    if(nums[j] == target){
                        res[1] = j;
                        return res;
                    }
                }
            }
            if(nums[mid]<target){
                lo = mid+1;
            }else{
                hi = mid-1;
            }
        }
        return res;
    }
}

方法二：

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = new int[]{-1,-1};
        int first = fun(nums,target-0.5);
        int last = fun(nums,target+0.5);
        if(first != last){
            res[0] = first;
            res[1] = last-1;
        }
        return res;
    }
    public int fun(int[] nums, double target){
        int lo = 0;
        int hi = nums.length-1;
        int mid = 0;
        while(lo<=hi){
            mid = lo + (hi-lo)/2;
            if(nums[mid]<target){
                lo = mid+1;
            }else{
                hi = mid-1;
            }
        }
        return lo;
    }
}