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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.

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

难度：easy

题目：

给定一个排序数组和目标值，如果目标值存在则返回目标值所在的位置，否则返回目标值所在的插入位置。

思路：二叉搜索树。

Runtime: 3 ms, faster than 59.63% of Java online submissions for Search Insert Position.

Memory Usage: 27.1 MB, less than 77.12% of Java online submissions for Search Insert Position.

class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (target == nums[mid]) {
                return mid;
            }

            if (target > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return left;
    }
}

Runtime: 3 ms, faster than 59.63% of Java online submissions for Search Insert Position.

Memory Usage: 28.7 MB, less than 12.43% of Java online submissions for Search Insert Position.

class Solution {
    public int searchInsert(int[] nums, int target) {
        return binarySearch(nums, 0, nums.length - 1, target);
    }
    
    private int binarySearch(int[] nums, int left, int right, int target) {
        if (left > right) {
            return left;
        }
        int mid = left + (right - left) / 2;
        if (target == nums[mid]) {
            return mid;
        }
        
        if (target > nums[mid]) {
            return binarySearch(nums, mid + 1, right, target);
        } else {
            return binarySearch(nums, left, mid - 1, target);
        }
    }
}

原文 https://segmentfault.com/a/1190000018094829

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