[?問題:?]

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.

翻譯：給你一個排好序的數組和一個目標值，請找出目標值能夠插入數組的位置。

[ 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：一定要考慮一些特殊情況，如數組為null等。

[ 解法:?]
①. 常規解法：從數組索引為0的位置開始找，時間復雜度為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 -> 2System.out.println(new Solution().searchInsert(arr, 2)); // 2 -> 1System.out.println(new Solution().searchInsert(arr, 7)); // 7 -> 4System.out.println(new Solution().searchInsert(arr, 0)); // 0 -> 0} }②. 二分查找：時間復雜度log2n 前提條件：一定是有序數組。

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 -> 2System.out.println(new Solution().searchInsert(arr, 2)); // 2 -> 1System.out.println(new Solution().searchInsert(arr, 7)); // 7 -> 4System.out.println(new Solution().searchInsert(arr, 0)); // 0 -> 0} }

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