【LeetCode】- Search Insert Position（查找插入的位置）

[ 问题: ]

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.

翻译：给你一个排好序的数组和一个目标值，请找出目标值可以插入数组的位置。

[ 分析: ]

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

注意：一定要考虑一些特殊情况，如数组为null等。

[ 解法: ]

①. 常规解法：从数组索引为0的位置开始找，时间复杂度为 O(n) ，accepted

public class Solution {
	public int searchInsert(int[] A, int target) {
		if (A != null) {
			for (int i = 0; i < A.length; i++) {
				if (target == A[i] || target < A[i]) {
					return i;
				} 
			}
			return A.length;
		}
		return -1;
	}

	public static void main(String[] args) {
		int[] arr = { 1, 3, 5, 6 };
		System.out.println(new Solution().searchInsert(arr, 5)); // 5 -> 2
		System.out.println(new Solution().searchInsert(arr, 2)); // 2 -> 1
		System.out.println(new Solution().searchInsert(arr, 7)); // 7 -> 4
		System.out.println(new Solution().searchInsert(arr, 0)); // 0 -> 0
	}
}

②. 二分查找：时间复杂度log2n

前提条件：一定是有序数组。

public class Solution {
	public int searchInsert(int[] A, int target) {
		int mid;
		int low = 0;
		int high = A.length - 1;
		while (low < high) {
			mid = (low + high) / 2;
			if (A[mid] < target) {
				low = mid + 1; 
			} else if (A[mid] > target) {
				high = mid - 1;
			} else {
				return mid;
			}
		}
		
		return target > A[low] ? low + 1 : low;
	}

	public static void main(String[] args) {
		int[] arr = { 1, 3, 5, 6 };
		System.out.println(new Solution().searchInsert(arr, 5)); // 5 -> 2
		System.out.println(new Solution().searchInsert(arr, 2)); // 2 -> 1
		System.out.println(new Solution().searchInsert(arr, 7)); // 7 -> 4
		System.out.println(new Solution().searchInsert(arr, 0)); // 0 -> 0
	}
}