Leetcode34 Search for a Range

Search for a Range

Given a sorted array of integers, 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution1

• 分为两步：
  
  • 按照正常的二分法找到一个target；
  • 找到target之后，分两头去确定前端点和后端点。
• 代码如下：

public class Solution {    public int[] searchRange(int[] nums, int target) {        int[] result = {-1,-1};        help(nums,0,nums.length-1,target,result);        return result;    }    public void help(int[] nums, int start, int end, int target, int[] result){        if(start>end) return;        int mid = (start+end)/2;        if(nums[mid]>target) help(nums,start,mid-1,target,result);//正常的二分        else if(nums[mid]<target) help(nums,mid+1,end,target,result);        else{//特殊在处理相等的情况            if(mid==0||nums[mid-1]!=target) result[0] = mid;//已经找到了前端所在位置            else help(nums,start,mid-1,target,result);//这里还不是前端，继续去前面找            if(mid==nums.length-1||nums[mid+1]!=target) result[1] = mid;            else help(nums,mid+1,end,target,result);        }    }}

Solution2

• 用迭代的方法稍微复杂些，主要原因在于这里要在某个点分两头去找，如果一直都只有一路二分搜索的话，迭代的方法非常简单，但是分两头必须得记录分叉口。可以如下实现：

public class Solution {    public int[] searchRange(int[] nums, int target) {        int[] result = {-1,-1};        int start = 0, end = nums.length-1, start2 = -1, end2 = -1;        while(start<=end){            int mid =(start+end)/2;            if(nums[mid]<target) start = mid+1;            else if(nums[mid]>target) end = mid-1;            else{                if(mid==nums.length-1||nums[mid+1]!=target) result[1] = mid;                else if(start2==-1){                    start2 = mid+1;                    end2 = end;                }                if(mid==0||nums[mid-1]!=nums[mid]){                    result[0] = mid;                    break;                }else end = mid-1;            }        }        if(start2==-1) return result;//数组中不存在target或者已经找到了前后端点        start = start2;        end = end2;        while(start<=end){            int mid =(start+end)/2;            if(nums[mid]<target) start = mid+1;            else if(nums[mid]>target) end = mid-1;            else{                if(mid==nums.length-1||nums[mid+1]!=nums[mid]){                    result[1] = mid;                    break;                }else start = mid+1;            }        }        return result;    }}

Solution3

• 将解法2进行稍微修改，如下：

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