35. Search Insert Position

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 4:

Input: [1,3,5,6], 0
Output: 0

解題思路：

首先想到的就是遍歷整個陣列，查詢與目標值相等的數值，然後返回對應的索引，

如果陣列中不存在目標值，那就返回剛好比目標值大的索引

如果目標值是最大的，那麼直接返回陣列的長度

此時的兩個if需要使用break跳出，否則會接著與下邊的數值進行比較

程式碼實現：

 1 package easy;
 2 
 3 class SolutionSearch {
 4     public int searchInsert(int
 
[] nums, int value) {
 5         int pos = 0;
 6         int i = 0;
 7         while(i < nums.length){
 8             if (nums[i] == value) {
 9                 pos = i;
10                 break;
11             } else {
12                 if (nums[i] > value) {
13                     pos =  i;
14
 
                     break;
15                 } else {
16                     pos =  nums.length;
17                 }
18             }
19             i ++;
20         }
21         return pos;
22     }
23 }
24 
25 public class SearchInsert {
26     public static void main(String[] args) {
27         SolutionSearch solution = new SolutionSearch();
28         int[] nums = {1,3,5,6};
29         int value = 2;
30         System.out.println(solution.searchInsert(nums, value));
31     }
32 }