LF.236.Search Insert Position
Given a sorted array and a target value, return the index where it would be if it were inserted in order.
Assumptions
If there are multiple elements with value same as target, we should insert the target before the first existing element.
Examples
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,3,3,5,6], 3 → 1
[1,3,5,6], 0 → 0
1 public int searchInsert(int[] nums, int target) {
2 //corner case
3 if (nums == null || nums.length ==0) {
4 return 0 ;
5 }
6 int left =0 , right = nums.length - 1 ;
7 if (nums[left] > target) {
8 return 0; // the first one
9 } else if(nums[right] < target){
10 return right +1 ; // the next one
11 }
12 //now its in the middle: if found, then return
13 int firstIndex = 0 ;
14 while(left + 1 < right){
15 int mid = left + (right - left)/2 ;
16 if (nums[mid] == target) {
17 right = mid ;
18 } else if(nums[mid] < target){
19 left = mid ;
20 } else {
21 right = mid ;
22 }
23 }
24 if (nums[left] == target) {
25 firstIndex = left ;
26 } else if (nums[right] == target) {
27 firstIndex = right ;
28 } else {
29 firstIndex = -1 ;
30 }
31 //the target will sit in the middle of the left and right
32 if (firstIndex == -1 ) {
33 return left+1;
34 } else{
35 return firstIndex ;
36 }
37 }
note, the leetcode one does not consider the duplicate scenario:
https://leetcode.com/problems/search-insert-position/description/
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Example 1:
Input: [1,3,5,6], 5
Output: 2
Example 2:
Input: [1,3,5,6], 2
Output: 1
Example 3:
Input: [1,3,5,6], 7
Output: 4
Example 1:
Input: [1,3,5,6], 0 Output: 0
1 public int searchInsert(int[] nums, int target) {
2 //corner case
3 if (nums == null || nums.length ==0) {
4 return 0 ;
5 }
6 int left =0 , right = nums.length - 1 ;
7 if (nums[left] >= target) {
8 return 0; // the first one
9 } else if(nums[right] == target){
10 return right ;
11 } else if(nums[right] < target){
12 return right +1 ; // the next one
13 }
14 //now its in the middle: if found, then return
15 while(left + 1 < right){
16 int mid = left + (right - left)/2 ;
17 if (nums[mid] == target) {
18 return mid ;
19 } else if(nums[mid] < target){
20 left = mid ;
21 } else {
22 right = mid ;
23 }
24 }
25 // post processing is required for the left + 1 < right way
26 if (nums[left] == target) {
27 return left ;
28 }
29 if (nums[right] == target) {
30 return right;
31 }
32 //the target will sit in the middle of the left and right
33 return left+1;
34 }
