35. Search Insert Position

Posted on In Words count: 999 Reading time ≈ 1 mins.

35. Search Insert Position

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Given a sorted array of distinct integers 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 must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [1,3,5,6], target = 5
Output: 2
Example 2:

Input: nums = [1,3,5,6], target = 2
Output: 1
Example 3:

Input: nums = [1,3,5,6], target = 7
Output: 4
Example 4:

Input: nums = [1,3,5,6], target = 0
Output: 0
Example 5:

Input: nums = [1], target = 0
Output: 0
 

Constraints:

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums contains distinct values sorted in ascending order.
-104 <= target <= 104

难度 : Easy

思路

这是用BinarySearch查找lowerbound的问题。
注意这里 r = n, 而且判断的条件是 nums[m] >= target.
因为搜索的范围是[l,r) 不包括r。这里每次缩小搜索范围的时候需要保证满足这个条件
结果就是l

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class Solution {
    public int searchInsert(int[] nums, int target) {
        
        int n = nums.length;
        int l = 0;
        int r = n;
        while(l < r) {
            int m = l + (r - l) / 2;
            if (nums[m] >= target) {
                r = m;
            } else {
                l = m + 1;
            }
        }
        return l;
    }
}