算法分析与设计——LeetCode:33. Search in Rotated Sorted Array

题目

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.

class Solution {
public:
    int search(vector<int>& nums, int target) {
    }
};

思路

使用二分查找，有三种情况：1.mid>begin>end；2.mid<end<begin；3.begin<mid<end，即正常排序的数列。 根据不同情况选择下一次查找的begin和end。

代码

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int end = nums.size();
        if (end == 0) {
            return -1;
        }
        int head = 0;
        int index = end/2;
        
        while ((end-head) != 0 && nums[index] != target) {
            if (nums[head] < nums[end-1]) {
                while ((end-head) != 0 && nums[index] != target) {
                    if (target > nums[index]) {
                        head = index+1;
                    } else {
                        end = index;
                    }
                    index = (end-head)/2+head;
                }
                break;
            }
            
            if(nums[index] > nums[head]) {
                if (target > nums[index]) {
                    head = index +1;
                } else {
                    if (target >= nums[head]) {
                        end = index;
                    } else {
                        head = index+1;
                    }
                }
            } else {
                if (target > nums[index]) {
                    if (target >= nums[head]) {
                        end = index;
                    } else {
                        head = index+1;
                    }
                } else {
                    end = index;
                }
            }
            index = (end-head)/2+head;
        }
        if (target != nums[index]) {
            return -1;
        } else {
            return index;
        }
    
    }
};