题目：
Given an array of integers 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].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
链接：https://leetcode.com/problems/search-for-a-range/#/description
4/14/2017
8ms, 53%
在找到第一个target之后，按照上行和下行分别找两个边界，代码可以refactor
1 public class Solution { 2 public int[] searchRange(int[] nums, int target) { 3 int[] ret = new int[2]; 4 ret[0] = -1; 5 ret[1] = -1; 6 if (nums.length == 0) return ret; 7 8 int lo = 0, hi = nums.length - 1; 9 int mid; 10 while (lo <= hi) { 11 mid = lo + (hi - lo) / 2; 12 if (nums[mid] == target) { 13 ret[0] = mid; 14 ret[1] = mid; 15 search(nums, target, mid + 1, hi, ret, false); 16 search(nums, target, lo, mid - 1, ret, true); 17 return ret; 18 } else if (nums[mid] < target) { 19 lo = mid + 1; 20 } else { 21 hi = mid - 1; 22 } 23 } 24 return ret; 25 } 26 27 private void search(int[] nums, int target, int lo, int hi, int[] ret, boolean down) { 28 int mid; 29 while (lo <= hi) { 30 mid = lo + (hi - lo) / 2; 31 if (nums[mid] == target) { 32 if (down) { 33 if (mid < ret[0]) { 34 ret[0] = mid; 35 } 36 hi = mid - 1; 37 } 38 else { 39 if (mid > ret[1]) { 40 ret[1] = mid; 41 } 42 lo = mid + 1; 43 } 44 } else if (nums[mid] < target) { 45 lo = mid + 1; 46 } else { 47 hi = mid - 1; 48 } 49 } 50 } 51 }
我看别人都是2次binarysearch分别搞定左边和右边，并且注意下标的变化第9，20行，并且在第一轮之后i并不需要重置到0
1 vector<int> searchRange(int A[], int n, int target) { 2 int i = 0, j = n - 1; 3 vector<int> ret(2, -1); 4 // Search for the left one 5 while (i < j) 6 { 7 int mid = (i + j) /2; 8 if (A[mid] < target) i = mid + 1; 9 else j = mid; 10 } 11 if (A[i]!=target) return ret; 12 else ret[0] = i; 13 14 // Search for the right one 15 j = n-1; // We don't have to set i to 0 the second time. 16 while (i < j) 17 { 18 int mid = (i + j) /2 + 1; // Make mid biased to the right 19 if (A[mid] > target) j = mid - 1; 20 else i = mid; // So that this won't make the search range stuck. 21 } 22 ret[1] = j; 23 return ret; 24 }
divide and conquer:
https://discuss.leetcode.com/topic/16486/9-11-lines-o-log-n
1 def searchRange(self, nums, target): 2 def search(lo, hi): 3 if nums[lo] == target == nums[hi]: 4 return [lo, hi] 5 if nums[lo] <= target <= nums[hi]: 6 mid = (lo + hi) / 2 7 l, r = search(lo, mid), search(mid+1, hi) 8 return max(l, r) if -1 in l+r else [l[0], r[1]] 9 return [-1, -1] 10 return search(0, len(nums)-1)
这个跟我的有点像，也比我好：
https://discuss.leetcode.com/topic/10692/simple-and-strict-o-logn-solution-in-java-using-recursion
更多讨论：
https://discuss.leetcode.com/category/42/search-for-a-range
4/27/2017
算法班
分别找左右边界
1 public class Solution { 2 /** 3 *@param A : an integer sorted array 4 *@param target : an integer to be inserted 5 *return : a list of length 2, [index1, index2] 6 */ 7 public int[] searchRange(int[] A, int target) { 8 // write your code here 9 int[] ret = new int[2]; 10 ret[0] = -1; 11 ret[1] = -1; 12 if (A == null || A.length == 0) { 13 return ret; 14 } 15 16 int start = 0, end = A.length - 1; 17 while (start + 1 < end) { 18 int mid = start + (end - start) / 2; 19 if (A[mid] >= target) { 20 end = mid; 21 } else { 22 start = mid; 23 } 24 } 25 if (A[start] == target) { 26 ret[0] = start; 27 } else if (A[end] == target) { 28 ret[0] = end; 29 } else { 30 return ret; 31 } 32 33 start = ret[0]; 34 end = A.length - 1; 35 while (start + 1 < end) { 36 int mid = start + (end - start) / 2; 37 if (A[mid] <= target) { 38 start = mid; 39 } else { 40 end = mid; 41 } 42 } 43 if (A[end] == target) { 44 ret[1] = end; 45 } else if (A[start] == target) { 46 ret[1] = start; 47 } else { 48 return ret; 49 } 50 51 return ret; 52 } 53 }