Leetcode Search in Rotated Sorted Array

最新推荐文章于 2024-05-03 20:14:24 发布

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最新推荐文章于 2024-05-03 20:14:24 发布

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分类专栏： OJ 文章标签： leetcode 二分搜索 oj lintcode

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本文介绍了一种在旋转排序数组中查找目标值的算法。通过两种方法实现：一是先找到最小元素的位置，然后分别进行二分搜索；二是直接在一个搜索过程中同时判断左右区间，避免了额外的最小元素查找步骤。

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Suppose a sorted array 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.

方法一：

先找最小数的下标，再分别二分搜索

//方法一：先找最小数的下标，再分别二分搜索
class Solution {
    /** 
     * param A : an integer ratated sorted array
     * param target :  an integer to be searched
     * return : an integer
     */
public:
    int searchPartition(vector<int> &A, int target, int left, int right){
        int l = left, r = right;
        if(l > r)   return -1;
        while(l < r){
            int mid = (l + r) / 2;
            if(A[mid] == target)    return mid;
            else if(A[mid] < target)    l = mid + 1;
            else    r = mid - 1;
        }
        if(A[l] == target)  return l;
        else    return -1;
    }

    int search(vector<int> &A, int target) {
        // write your code here
        int n = A.size();
        if(n == 0)  return -1;
        int left = 0, right = n - 1;
        int minIndex;
        while(left + 1 < right && A[left] > A[right]){
            int mid = (left + right) / 2;
            if(A[mid] < A[right])  right = mid;
            else if(A[mid] > A[left])   left = mid;
        }
        minIndex = A[left] < A[right] ? left : right;
        int findLeft = searchPartition(A, target, 0, minIndex - 1);
        int findRight = searchPartition(A, target, minIndex, n - 1);
        if(findLeft == -1){
            if(findRight == -1)  return -1;
            else    return findRight;
        }
        else    return findLeft;
    }
};

方法二：

同时考虑

//方法二：一起考虑
class Solution {
    /** 
     * param A : an integer ratated sorted array
     * param target :  an integer to be searched
     * return : an integer
     */
public:
   int search(vector<int> &A, int target) {
        // write your code here
        int n = A.size();
        if(n == 0)  return -1;
        int left = 0, right = n - 1;
        while(left < right){
            int mid = (left + right) / 2;
            if(A[mid] == target)    return mid;
            if(A[mid] > A[left]){
                if(target < A[mid] && target >= A[left])    //注意要>=
                    right = mid - 1;
                else    left = mid + 1; //即target<A[mid]也<A[left]和target>A[mid]
            }
            else{
                if(target > A[mid] && target <= A[right]) 
                    left = mid + 1;
                else    right = mid - 1;
            }
        }
        if(A[left] == target)   return left;
        else    return -1;
    }
};