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| 1 | +class Solution { |
| 2 | + /* |
| 3 | + Rule 1 |
| 4 | + - Given array is sorted in non-decreasing order |
| 5 | + |
| 6 | + Rule 2 |
| 7 | + - Limit time complexity O(log n). |
| 8 | + |
| 9 | + So I can use Binary Search Algorithm. |
| 10 | + */ |
| 11 | + public int[] searchRange(int[] nums, int target) { |
| 12 | + // initialize arr[0] -> maximum integer, arr[1] -> minimum integer |
| 13 | + int[] arr = {100001, -10}; |
| 14 | + int s = 0; |
| 15 | + int e = nums.length - 1; |
| 16 | + |
| 17 | + // find minimum index in nums with Binary Search alogorithm |
| 18 | + while (s <= e) { |
| 19 | + |
| 20 | + int mid = (s + e) / 2; |
| 21 | + |
| 22 | + if(nums[mid] > target) { |
| 23 | + e = mid - 1; |
| 24 | + } |
| 25 | + else if(nums[mid] <= target) { |
| 26 | + if(nums[mid] == target) { |
| 27 | + arr[0] = Math.min(arr[0], mid); |
| 28 | + arr[1] = Math.max(arr[1], mid); |
| 29 | + } |
| 30 | + s = mid + 1; |
| 31 | + } |
| 32 | + } |
| 33 | + |
| 34 | + s = 0; |
| 35 | + e = nums.length - 1; |
| 36 | + |
| 37 | + // find maximum index in nums with Binary Search alogorithm |
| 38 | + while(s <= e) { |
| 39 | + int mid = (s + e) / 2; |
| 40 | + |
| 41 | + if(nums[mid] >= target) { |
| 42 | + if(nums[mid] == target) { |
| 43 | + arr[0] = Math.min(arr[0], mid); |
| 44 | + arr[1] = Math.max(arr[1], mid); |
| 45 | + } |
| 46 | + e = mid - 1; |
| 47 | + } |
| 48 | + else if(nums[mid] < target) { |
| 49 | + s = mid + 1; |
| 50 | + } |
| 51 | + } |
| 52 | + |
| 53 | + // if arr data is initial data, set -1. |
| 54 | + if(arr[0] == 100001) arr[0] = -1; |
| 55 | + if(arr[1] == -10) arr[1] = -1; |
| 56 | + |
| 57 | + return arr; |
| 58 | + } |
| 59 | +} |
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