Leetcode 34. Search for a Range

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]

.

For example,
Given

[5, 7, 7, 8, 8, 10]

and target value 8,
return

[3, 4]

.

解题思路：

题意转换， 变成找first bound of target and right bound of target.

找first position:

相同的时候砍掉右边

找last position:

相同的时候砍掉左边

题目可以变化问题：问8（target number）出现了几次？

答案：right bound - left bound + 1

Java code:

public class Solution {
public int[] searchRange(int[] nums, int target) {
if(nums.length == 0 ){
return new int[]{-1, -1};
}

int start, end, mid;
int[] bound = new int[2];

//search for left bound
start = 0;
end = nums.length-1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (nums[mid] == target) {
end = mid;
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if(nums[start] == target){
bound[0] = start;
} else if (nums[end] == target) {
bound[0] = end;
} else {
bound[0] = bound[1] = -1;
return bound;
}

//search for right bound
start = 0;
end = nums.length-1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (nums[mid] == target) {
start = mid;
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if(nums[end] == target) {
bound[1] = end;
}else if (nums[start] == target){
bound[1] = start;
}else {
bound[0] = bound[1] = -1;
return bound;
}

return bound;
}
}

Reference:

1. http://www.jiuzhang.com/solutions/search-for-a-range/