leetcode -- Search for a Range
添加时间:2013-8-1 点击量:
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithms 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]
.
思路:
应用二分搜刮找到target的idx,然后查看该idx的阁下断定局限。
算法错杂度:
均匀景象下是O(lgn);
最坏景象下数组中所有元素都雷同O(n);
1 public class Solution {
2 public int[] searchRange(int[] A, int target) {
3 // Start typing your Java solution below
4 // DO NOT write main() function
5 int idx = binarySearch(A, target);
6 int len = A.length;
7 int[] results = null;
8 if(idx == -1){
9 results = new int[]{-1, -1};
10 } else{
11 int l = idx;
12 int r = idx;
13 while(l >= 0 && A[l] == target){
14 l--;
15 }
16 l++;
17
18 while(r < len && A[r] == target){
19 r++;
20 }
21 r--;
22 results = new int[]{l, r};
23 }
24 return results;
25 }
26
27 public int binarySearch(int[] A, int target){
28 int len = A.length;
29 int l = 0, r = len - 1;
30 while(l <= r){
31 int mid = (l + r) / 2;
32 if(target == A[mid])
33 return mid;
34
35 if(target > A[mid]){
36 l = mid + 1;
37 } else {
38 r = mid - 1;
39 }
40 }
41
42 return -1;
43 }
44 }
所有随风而逝的都属于昨天的,所有历经风雨留下来的才是面向未来的。—— 玛格丽特·米切尔 《飘》
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithms 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]
.
思路:
应用二分搜刮找到target的idx,然后查看该idx的阁下断定局限。
算法错杂度:
均匀景象下是O(lgn);
最坏景象下数组中所有元素都雷同O(n);
1 public class Solution {
2 public int[] searchRange(int[] A, int target) {
3 // Start typing your Java solution below
4 // DO NOT write main() function
5 int idx = binarySearch(A, target);
6 int len = A.length;
7 int[] results = null;
8 if(idx == -1){
9 results = new int[]{-1, -1};
10 } else{
11 int l = idx;
12 int r = idx;
13 while(l >= 0 && A[l] == target){
14 l--;
15 }
16 l++;
17
18 while(r < len && A[r] == target){
19 r++;
20 }
21 r--;
22 results = new int[]{l, r};
23 }
24 return results;
25 }
26
27 public int binarySearch(int[] A, int target){
28 int len = A.length;
29 int l = 0, r = len - 1;
30 while(l <= r){
31 int mid = (l + r) / 2;
32 if(target == A[mid])
33 return mid;
34
35 if(target > A[mid]){
36 l = mid + 1;
37 } else {
38 r = mid - 1;
39 }
40 }
41
42 return -1;
43 }
44 }
所有随风而逝的都属于昨天的,所有历经风雨留下来的才是面向未来的。—— 玛格丽特·米切尔 《飘》