C++实现LeetCode(81.在旋转有序数组中搜索之二)

[LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索之二

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

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

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

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

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

  • This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
  • Would this affect the run-time complexity? How and why?

这道是之前那道 Search in Rotated Sorted Array 的延伸，现在数组中允许出现重复数字，这个也会影响我们选择哪半边继续搜索，由于之前那道题不存在相同值，我们在比较中间值和最右值时就完全符合之前所说的规律：如果中间的数小于最右边的数，则右半段是有序的，若中间数大于最右边数，则左半段是有序的。而如果可以有重复值，就会出现来面两种情况，[3 1 1] 和 [1 1 3 1]，对于这两种情况中间值等于最右值时，目标值3既可以在左边又可以在右边，那怎么办么，对于这种情况其实处理非常简单，只要把最右值向左一位即可继续循环，如果还相同则继续移，直到移到不同值为止，然后其他部分还采用 Search in Rotated Sorted Array 中的方法，可以得到代码如下：

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        int n = nums.size(), left = 0, right = n - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) return true;
            if (nums[mid] < nums[right]) {
                if (nums[mid] < target && nums[right] >= target) left = mid + 1;
                else right = mid - 1;
            } else if (nums[mid] > nums[right]){
                if (nums[left] <= target && nums[mid] > target) right = mid - 1;
                else left = mid + 1;
            } else --right;
        }
        return false;
    }
};

到此这篇关于C++实现LeetCode(81.在旋转有序数组中搜索之二)的文章就介绍到这了,更多相关C++实现在旋转有序数组中搜索之二内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(73.矩阵赋零)

    [LeetCode] 73.Set Matrix Zeroes 矩阵赋零 Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. click to show follow up. Follow up: Did you use extra space? A straight forward solution using O(mn) space is probably

  • C++实现LeetCode(79.词语搜索)

    [LeetCode] 79. Word Search 词语搜索 Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. Th

  • C++实现LeetCode(76.最小窗口子串)

    [LeetCode] 76. Minimum Window Substring 最小窗口子串 Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BA

  • C++实现LeetCode(80.有序数组中去除重复项之二)

    [LeetCode] 80. Remove Duplicates from Sorted Array II 有序数组中去除重复项之二 Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you mus

  • C++实现LeetCode(74.搜索一个二维矩阵)

    C++实现LeetCode(74.搜索一个二维矩阵)

    [LeetCode] 74. Search a 2D Matrix 搜索一个二维矩阵 Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right. The first integer of each row is great

  • C++实现LeetCode(82.移除有序链表中的重复项之二)

    [LeetCode] 82. Remove Duplicates from Sorted List II 移除有序链表中的重复项之二 Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Example 1: Input: 1->2->3->3->4->4->5 Outp

  • C++实现LeetCode(75.颜色排序)

    [LeetCode] 75. Sort Colors 颜色排序 Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2

  • C++实现LeetCode(72.编辑距离)

    [LeetCode] 72. Edit Distance 编辑距离 Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a char

  • C++实现LeetCode(81.在旋转有序数组中搜索之二)

    [LeetCode] 81. Search in Rotated Sorted Array II 在旋转有序数组中搜索之二 Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search.

  • C++实现LeetCode(33.在旋转有序数组中搜索)

    [LeetCode] 33. Search in Rotated Sorted Array 在旋转有序数组中搜索 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 f

  • C++实现LeetCode(154.寻找旋转有序数组的最小值之二)

    [LeetCode] 154. Find Minimum in Rotated Sorted Array II 寻找旋转有序数组的最小值之二 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]). Find the minimum element. Th

  • C++实现LeetCode(153.寻找旋转有序数组的最小值)

    [LeetCode] 153. Find Minimum in Rotated Sorted Array 寻找旋转有序数组的最小值 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]). Find the minimum element. You may

  • C++实现LeetCode(88.混合插入有序数组)

    [LeetCode] 88. Merge Sorted Array 混合插入有序数组 Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: The number of elements initialized in nums1and nums2 are m and n respectively. You may assume that nums1 has

  • C++实现LeetCode(26.有序数组中去除重复项)

    [LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项 Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this

  • C++实现LeetCode(34.在有序数组中查找元素的第一个和最后一个位置)

    [LeetCode] 34. Find First and Last Position of Element in Sorted Array 在有序数组中查找元素的第一个和最后一个位置 Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity

  • java算法入门之有效的括号删除有序数组中的重复项实现strStr

    java算法入门之有效的括号删除有序数组中的重复项实现strStr

    目录 1.LeetCode 20.有效的括号 题目 小编菜解 思路及算法 大神解法 2.LeetCode 26.删除有序数组中的重复项 题目 小编菜解初版 小编菜解改进版 思路及算法 大神解法 3.LeetCode 28.实现strStr 题目 小编菜解 大神解法 也许,我们永远都不会知道自己能走到何方,遇见何人,最后会变成什么样的人,但一定要记住,能让自己登高的,永远不是别人的肩膀,而是挑灯夜战的自己,人生的道路刚刚启程,当你累了倦了也不要迷茫,回头看一看,你早已不再是那个年少轻狂的少年. 1

  • java实现向有序数组中插入一个元素实例

    整理文档,搜刮出一个java实现向有序数组中插入一个元素,稍微整理精简一下做下分享 package cn.jbit.array; import java.util.*; public class Insert { public static void main(String[] args) { //字符排序 char[] chars = new char[9]; chars[0] = 'a'; chars[1] = 'c'; chars[2] = 'u'; chars[3] = 'b'; cha

随机推荐