【數組】B054_搜索旋轉排序數組（二分找單調區間）

一、Problem

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).

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

二、Solution

方法一：二分

• 樸素的二分搜索要求數組是單調的，而經過旋轉的數組必定具有兩種單調性。所以如何找到單調區間成爲本題重點，本題可劃分爲幾步：
  • 先找到一個連續遞增區間。
  • 從遞增區間裏面再次確定左右邊界。

邊界錯誤：這裏需要注意的是，當出現等於 = 的情況，我們依舊要把邊界移動，否則，會進入死循環…😂

public int search(int[] A, int t) {
    int l = 0, r = A.length-1;
    while (l <= r) {
        int m = (l + r) >>> 1;
        if (A[m] == t)
            return m;
        if (A[l] <= A[m]) {
            if (A[l] <= t && t <= A[m]) r = m - 1;
            else l = m + 1;
        } else {
            if (t >= A[m] && t <= A[r]) l = m + 1;
            else r = m;
        }
    }
    return -1;
}

複雜度分析

• 時間複雜度：$O(logN)$，
• 空間複雜度：$O(1)$，