Search for a Range
1.最简单的想法，用最普通的二分查找，找到target,然后向左右扩张，大量的重复的target，就会出现O(n)效率。
class Solution { public int[] searchRange(int[] A, int target) { int ans[]=new int[2]; int a= bserch(A,target); if(a==-1) {ans[0]=-1;ans[1]=-1; return ans;} //left int b=a-1; while(b>=0&&A[b]==target) b--; ans[0]=b+1; b=a+1; while(b<=A.length-1&&A[b]==target) b++; ans[1]=b-1; return ans; } public int bserch(int A[],int target) { int low=0; int end=A.length-1; while(low<=end) { int mid=(low+end)>>1; if(A[mid]==target) return mid; else if(A[mid]<target) low=mid+1; else end=mid-1; } return -1; } }
2.二分查找加入第一个大的位置（上届），第一个大于等于它的位置（下界）
1 class Solution { 2 public int[] searchRange(int[] A, int target) { 3 int ans[]=new int[2]; 4 int a= bserch(A,target); 5 if(a==-1) {ans[0]=-1;ans[1]=-1; return ans;} 6 //left 7 int b=a-1; 8 while(b>=0&&A[b]==target) b--; 9 ans[0]=b+1; 10 b=a+1; 11 12 while(b<=A.length-1&&A[b]==target) b++; 13 ans[1]=b-1; 14 return ans; 15 16 17 18 19 } 20 //其实难点就是A[mid]==target，如何调整low和high 21 public int bserch(int A[],int target) 22 { 23 int low=0; 24 int end=A.length-1; 25 while(low<=end) 26 { 27 int mid=(low+end)>>1; 28 if(A[mid]==target) return mid; 29 else if(A[mid]<target) low=mid+1; 30 else end=mid-1; 31 32 } 33 34 return -1; 35 }
//第一个大于target的 36 public int upperbound(int A[],int target) // the first one larger than target 37 { 38 int low=0; 39 int end=A.length-1; 40 while(low<=end) 41 { 42 int mid=(low+end)>>1; 43 if(A[mid]>target) end=mid-1; 44 else low=mid+1; 45 46 } 47 return low; 48 } 49 public int lowbound(int A[],int target) // the first one larger or equal target 50 { 51 int low=0; 52 int end=A.length-1; 53 while(low<=end) 54 { 55 int mid=(low+end)>>1; 56 if(A[mid]>=target) end=mid-1; 57 else low=mid+1; 58 59 } 60 return low; 61 } 62 63 64 65 66 67 }