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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].
Analysis: 这道题是二分查找Search Insert Position的变体，思路并不复杂，就是先用二分查找找到其中一个target，然后再往左右找到target的边缘。找边缘的方法跟二分查找仍然是一样的，只是相等的情况依然要往左找（找左边缘）或往右找（找右边缘）。这样下来总共进行了三次二分查找，所以算法的时间复杂度仍是O(logn)，空间复杂度是O(1)。
Notice: 对停止的边界条件极不熟悉，需要总结，参见Binary Search的Summary
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int start = 0, end = A.length - 1; 4 int[] result = new int[2]; 5 result[0] = -1; 6 result[1] = -1; 7 if(A==null || A.length==0) 8 { 9 return result; 10 } 11 int middle = search(A, start, end, target, result); 12 if (middle == -1) return result; 13 result[0] = searchleft(A, 0, middle, target); 14 result[1] = searchright(A, middle, A.length-1, target); 15 return result; 16 } 17 18 public int search(int[] A, int start, int end, int target, int[] result) { 19 int mid = -1; 20 if (start <= end) { 21 mid = (start + end) / 2; 22 if (A[mid] == target) { 23 result[0] = mid; 24 result[1] = mid; 25 return mid; 26 } 27 else if (A[mid] < target) { 28 start = mid + 1; 29 return search(A, start, end, target, result); 30 } 31 else { 32 end = mid - 1; 33 return search(A, start, end, target, result); 34 } 35 } 36 else { 37 result[0] = -1; 38 result[1] = -1; 39 return mid; 40 } 41 42 } 43 44 public int searchleft(int[] A, int start, int end, int target) { 45 if (start > end) return start; 46 int mid = (start + end) / 2; 47 if (A[mid] < target) { 48 start = mid + 1; 49 return searchleft(A, start, end, target); 50 } 51 else { 52 end = mid - 1; 53 return searchleft(A, start, end, target); 54 } 55 } 56 57 public int searchright(int[] A, int start, int end, int target) { 58 if (start > end) return end; 59 int mid = (start + end) / 2; 60 if (A[mid] <= target) { 61 start = mid + 1; 62 return searchright(A, start, end, target); 63 } 64 else { 65 end = mid - 1; 66 return searchright(A, start, end, target); 67 } 68 } 69 }
如果找到target元素之后寻找边界，搜索范围过大，可以把搜索范围缩小到找到target之前最近一次(start, end)区间内，如下：
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int start = 0, end = A.length - 1; 4 int[] result = new int[2]; 5 result[0] = -1; 6 result[1] = -1; 7 if(A==null || A.length==0) 8 { 9 return result; 10 } 11 int[] range = new int[2]; 12 int middle = search(A, start, end, target, range); 13 if (middle == -1) return result; 14 result[0] = searchleft(A, range[0], middle, target); 15 result[1] = searchright(A, middle, range[1], target); 16 return result; 17 } 18 19 public int search(int[] A, int start, int end, int target, int[] range) { 20 range[0] = start; 21 range[1] = end; 22 int mid = -1; 23 if (start <= end) { 24 mid = (start + end) / 2; 25 if (A[mid] == target) { 26 return mid; 27 } 28 else if (A[mid] < target) { 29 start = mid + 1; 30 return search(A, start, end, target, range); 31 } 32 else { 33 end = mid - 1; 34 return search(A, start, end, target, range); 35 } 36 } 37 else { 38 return mid; 39 } 40 41 } 42 43 public int searchleft(int[] A, int start, int end, int target) { 44 if (start > end) return start; 45 int mid = (start + end) / 2; 46 if (A[mid] < target) { 47 start = mid + 1; 48 return searchleft(A, start, end, target); 49 } 50 else { 51 end = mid - 1; 52 return searchleft(A, start, end, target); 53 } 54 } 55 56 public int searchright(int[] A, int start, int end, int target) { 57 if (start > end) return end; 58 int mid = (start + end) / 2; 59 if (A[mid] <= target) { 60 start = mid + 1; 61 return searchright(A, start, end, target); 62 } 63 else { 64 end = mid - 1; 65 return searchright(A, start, end, target); 66 } 67 } 68 }
如果想用iterative的方法来写：
1 public class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int[] res = new int[2]; 4 res[0] = -1; 5 res[1] = -1; 6 if(A==null || A.length==0) 7 { 8 return res; 9 } 10 int l=0; 11 int r=A.length-1; 12 int m=(l+r)/2; 13 while(l<=r) 14 { 15 m=(l+r)/2; 16 if(A[m]==target) 17 { 18 res[0]=m; 19 res[1]=m; 20 break; 21 } 22 else if(A[m]>target) 23 { 24 r = m-1; 25 } 26 else 27 { 28 l = m+1; 29 } 30 } 31 if(A[m]!=target) 32 return res; 33 int newL = m; 34 int newR = A.length-1; 35 while(newL<=newR) 36 { 37 int newM=(newL+newR)/2; 38 if(A[newM]==target) 39 { 40 newL = newM+1; 41 } 42 else 43 { 44 newR = newM-1; 45 } 46 } 47 res[1]=newR; 48 newL = 0; 49 newR = m; 50 while(newL<=newR) 51 { 52 int newM=(newL+newR)/2; 53 if(A[newM]==target) 54 { 55 newR = newM-1; 56 } 57 else 58 { 59 newL = newM+1; 60 } 61 } 62 res[0]=newL; 63 64 return res; 65 66 } 67 }
Leetcode: Search for a Range,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/EdwardLiu/p/3811193.html
