Difficulty: Medium
Given an array of integers nums sorted in ascending order, 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 1:
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| Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
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Example 2:
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| Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
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Solution
Language: Java
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| class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return new int[] {-1, -1};
}
int left, right = 0;
int start = 0;
int end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
start = mid;
} else if (nums[mid] > target) {
end = mid;
} else {
start = mid;
}
}
if (nums[end] == target) {
right = end;
} else if (nums[start] == target) {
right = start;
} else {
return new int[] {-1, -1};
}
start = 0;
end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
end = mid;
} else if (nums[mid] > target) {
end = mid;
} else {
start = mid;
}
}
if (nums[start] == target) {
left = start;
} else {
left = end;
}
return new int[] {left, right};
}
}
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| class Solution {
public int[] searchRange(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return new int[]{-1, -1};
}
return new int[] {search(nums, target, true), search(nums, target, false)};
}
private int search(int[] nums, int target, boolean isLeft) {
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
if (isLeft) {
end = mid;
} else {
start = mid;
}
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (isLeft) {
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
} else {
if (nums[end] == target) {
return end;
}
if (nums[start] == target) {
return start;
}
}
return -1;
}
}
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