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·186 lines (152 loc) · 4.25 KB
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M
1528173258
tags: Array, Binary Search
给sorted array, 有重复数字, 找跟target重合所在的range.
#### Binary Search
- 2个while loop
- 找first/last occurance
- TODO: Can the code be simplified?
```
/*
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].
Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
Tags Expand
Binary Search Array Sorted Array
*/
/*
Recap: 12.08.2015
input sorted
2 binary search while loop.
First one, keep looking for 1st occurnace.
Second one, keep looking for alst occurance.
check border case:
A = []
*/
public class Solution {
public int[] searchRange(int[] A, int target) {
int[] rst = new int[]{-1, -1};
if (A == null || A.length == 0) {
return rst;
}
int start = 0;
int end = A.length - 1;
int mid = start + (end - start)/2;
//1st occurance.
int first = 0;
while (start + 1 < end) {
mid = start + (end - start)/2;
if (A[mid] == target) {
if (mid - 1 >= 0 && A[mid - 1] == target) {
end = mid;
continue;
}
break;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (A[start] == target) {
first = start;
} else if (A[mid] == target) {
first = mid;
} else if (A[end] == target) {
first = end;
} else {
return rst;
}
//check last occurance
int last = first;
start = first;
end = A.length - 1;
while (start + 1 < end) {
mid = start + (end - start)/2;
if (A[mid] == target) {
if (mid + 1 < A.length && A[mid + 1] == target) {
start = mid;
continue;
}
break;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (A[end] == target) {
last = end;
} else if (A[mid] == target) {
last = mid;
} else if (A[start] == target) {
last = start;
}
rst[0] = first;
rst[1] = last;
return rst;
}
}
//Older solution:
public class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public ArrayList<Integer> searchRange(ArrayList<Integer> A, int target) {
// write your code here
int start = 0;
int end = A.size() - 1;
int mid;
ArrayList<Integer> bound = new ArrayList<Integer>();
bound.add(-1);
bound.add(-1);
if ( A.size() == 0) {
return bound;
}
//Left:
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A.get(mid) < target) {
start = mid;
} else if (A.get(mid) > target) {
end = mid;
} else {
end = mid;//Run to left
}
}
if (A.get(start) == target) {//Run to left
bound.set(0, start);
} else if (A.get(end) == target) {
bound.set(0, end);
} else {
return bound;
}
//Right:
start = 0;
end = A.size() - 1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A.get(mid) < target) {
start = mid;
} else if (A.get(mid) > target) {
end = mid;
} else {
start = mid;//Run to right
}
}
if (A.get(end) == target) {//Run to right
bound.set(1, end);
} else if (A.get(start) == target) {
bound.set(1, start);
} else {
return bound;
}
return bound;
}
}
```