Leetcode_35_Search Insert Position

本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/43739647

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

思路：

（1）题意为给定一个已排序的数组和一个整数，求该整数在数组中的位置。

（2）由于数组是已排序的，本题可以通过“二分查找”算法来寻找目标整数的位置。但需要注意对边界值判断，当目标整数小于数组第一个元素时，应返回0；当目标整数大于数组最后一个元素时，返回数组长度加1。该题还是比较简单，详情见下方代码。

（3）希望本文对你有所帮助。

算法代码实现如下：

/*** @author liqq*/
public class SearchInsert {public int searchInsert(int[] A, int target) {int high = A.length - 1;int low = 0;int mid = (high + low) / 2;if (A[low] > target)return 0;if (A[high] < target)return high + 1;while (mid >= 0 || mid <= high) {if (A[mid] > target) {if (A[mid] > target && A[mid - 1] < target) {return mid;} else {mid--;}} else if (A[mid] < target) {if (A[mid] < target && A[mid + 1] > target) {return mid + 1;} else {mid++;}} else {return mid;}}return -1;}
}

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