35. Search Insert Position

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.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

题意：给定一个排序的数组和一个目标值，如果目标值在数组里返回其下标，如果不在，返回其对应的插入位置。

思路：二分查找。

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1. class Solution {  
2. public:  
3.     int searchInsert(vector<int>& nums, int target) {  
4.         if (nums.empty())  
5.             return 0;  
6.         return binarySearch(nums, 0, nums.size() - 1, target);  
7.     }  
8.     int binarySearch(vector<int>& nums, int left, int right, int target){  
9.         if (left == right){  
10.             if (nums[left] < target)  
11.                 return left + 1;  
12.             if (nums[left] >= target)  
13.                 return left;  
14.         }  
15.         int mid = (left + right) / 2;  
16.         if (nums[mid] == target)  
17.             return mid;  
18.         else if (nums[mid] > target){  
19.             int newRight = mid - 1;  
20.             if (newRight < left)  
21.                 return left;  
22.             return binarySearch(nums, left, newRight, target);  
23.         }  
24.         else{  
25.             int newLeft = mid + 1;  
26.             if (newLeft > right)  
27.                 return right + 1;  
28.             return binarySearch(nums, newLeft, right, target);  
29.         }  
30.     }  
31. };