last position of target / first position of target

Last Position of Target

Find the last position of a target number in a sorted array. Return -1 if target does not exist.

O(log n)

public class Solution {
/**
* @param nums: An integer array sorted in ascending order
* @param target: An integer
* @return an integer
*/
public int lastPosition(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0;
int end = nums.length - 1;
while (start + 1 < end) {
int mid = (start + end) / 2;
if (target < nums[mid]) { //向后逼近
end = mid;
} else {
start = mid;
}
}
if (nums[end] == target) {
return end;
} else if (nums[start] == target) {
return start;
} else {
return -1;
}
}
}

First Position of Target

For a given sorted array (ascending order) and a 

target

number, find
the first index of this number in 

O(log n)

time complexity.

If the target number does not exist in the array, return 

-1

.

Example

If the array is 

[1,
2, 3, 3, 4, 5, 10]

, for given target 

3

, return 

2

.
O(log n)

class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
if(nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if(nums[mid] < target) { // 向前逼近
start = mid;
} else {
end = mid;
}
}
if(nums[start] == target) {//check the first position first
return start;
} else if(nums[end] == target) {
return end;
}
return -1;
}
}