首页 > 编程语言 > 详细

Java for LeetCode 035 Search Insert Position

时间:2015-05-08 01:39:25      阅读:284      评论:0      收藏:0      [点我收藏+]

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

解题思路一:

二分查找,注意边界条件,JAVA实现如下:

	static public int searchInsert(int[] nums, int target) {
		int left = 0, right = nums.length - 1;
		while (left <= right) {
			if (target == nums[(left + right) / 2])
				return (left + right) / 2;
			else if (target < nums[(left + right) / 2]) {
				if ((left + right) / 2 - 1 < left)
					return target > nums[left] ? left + 1 : left;
				if (target > nums[(left + right) / 2 - 1])
					return (left + right) / 2;
				if (target == nums[(left + right) / 2 - 1])
					return (left + right) / 2 - 1;
				right = (left + right) / 2 - 1;

			} else {
				if ((left + right) / 2 + 1 > right)
					return target > nums[right] ? right + 1 : right;
				if (target <= nums[(left + right) / 2 + 1])
					return (left + right) / 2 + 1;
				left = (left + right) / 2 + 1;
			}
		}
		return right;
	}

解题思路二:

同样采用二分,可以通过某些方法,减少代码量,JAVA实现如下:

static public int searchInsert(int[] nums, int target) {
		int left = 0, right = nums.length - 1;  
        while (left < right) {    
            if (nums[(right + left) / 2] < target)  
            	left = (right + left) / 2 + 1;  
            else  
            	right =  (right + left) / 2;  
        }  
        return nums[left] >= target ? left : left + 1;  
	}

 

Java for LeetCode 035 Search Insert Position

原文:http://www.cnblogs.com/tonyluis/p/4486580.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!