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  • 旋转已排序数组中查找

    1. 数组中无重复元素

    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

    1.1 寻找最小值所在的点
    建立模型求解

    本题关键在于求解 最小值所在的索引,再通过二分法求解即可。

    选择right作为比较的轴值,原因在于nums[right] 永远不会等于nums[mid],分一下三种情况讨论。

    如果中间值比最右端的值大,那么应该让 left = mid + 1,如果中间值比最右端小,那么应该让right = mid,因为nums[mid] = y1时也是满足中间值比最右端小,不应该让right = mid - 1.

    循环结束时left = right = index(y1)

     1.2 二分法求解

    先按照标准的二分法思路求解,不同的是,旋转以后,中间值与未旋转之前总是向后偏移最小值索引个单位。

     int search(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        int mid = 0;
    
        while (left < right) {
            mid = (left + right) / 2;
            if (nums[mid] <= nums[right]) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        int point = left;
        left = 0;
        right = nums.size() - 1;
        while (left <= right) {
            int medium = (left + right) / 2;
            mid = (medium + point) % nums.size();
            if (nums[mid] < target) {
                left = medium + 1;
            }
            else if (nums[mid] > target) {
                right = medium - 1;
            }
            else {
                return mid;
            }
        }
        return -1;
    }
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  • 原文地址:https://www.cnblogs.com/feng-ying/p/11225475.html
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