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  • Search in Rotated Sorted Array——LeetCode

    Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

    (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

    You are given a target value to search. If found in the array return its index, otherwise return -1.

    You may assume no duplicate exists in the array.

    Your algorithm's runtime complexity must be in the order of O(log n).

    Example 1:

    Input: nums = [4,5,6,7,0,1,2], target = 0
    Output: 4
    

    Example 2:

    Input: nums = [4,5,6,7,0,1,2], target = 3
    Output: -1

    题目大意:有序数组旋转,寻找其中是否有某个特定的值,时间复杂度O(log n)。

    思路:旋转后的数组,旋转后的后一部分肯定比前一部分所有元素都小;所以数组从中间分开,至少有一半是有序的。

    那么可以得到,如果中间的元素(也就是nums[mid])比nums[0]大,那么前半部分有序,如果中间的元素比nums[len-1]小,那么后半部分有序,这两种情况只会出现一种。这样就可以先看这个数在不在有序的这一半数组中,在的话没什么好说的,二分查找;不在的话,继续找另一半RotatedArray。思路就是这么个思路,然而想把题目A掉还要考虑很多细节,尤其是边界值,> or >= ,< or <=等等,极易出错。

    public int search(int[] nums, int target) {
            if (nums == null || nums.length == 0) {
                return -1;
            }
            int low = 0, high = nums.length - 1;
            while (low <= high) {
                int mid = (low + high) >> 1;
                if (nums[mid] == target) {
                    return mid;
                }
                if (nums[low] <= nums[mid]) {
                    if (target >= nums[low] && target <= nums[mid]) {
                        high = mid - 1;
                    } else {
                        low = mid + 1;
                    }
                } else {
                    if (target >= nums[mid] && target <= nums[high]) {
                        low = mid + 1;
                    } else {
                        high = mid - 1;
                    }
                }
            }
            return -1;
        }
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  • 原文地址:https://www.cnblogs.com/aboutblank/p/9644039.html
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