LeetCode-Search in Rotated Sorted Array(在反转数组中的关键字)

向右看齐 2022-08-07 03:55 2阅读 0赞

题1:

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.

题2：如果数组有重复数字呢？

对于此题，可以先解决[此题][Link 1]，找出反转数组中最小值。有了这题的基础，这题就简单了。

public int search(int[] A, int target) {
        	int low = 0;
            int high = A.length-1;
    		while (low <= high) {
    			int mid = low + (high - low) / 2;
    			if (A[mid] == target) {
    				return mid;
    			} else if (A[mid] < A[high]) {//右侧有序
    				if (A[mid] <= target && target <= A[high]) {
    					low = mid+1;
    				} else {
    					high = mid-1;
    				}
    			} else {//左侧有序
    				if (A[low] <= target && target <= A[mid]) {
    					high = mid-1;
    				} else {
    					low = mid+1;
    				}
    			}
    		}
            return -1;
        }

因为可以根据（low,mid）或（mid,high）有序，解决之。题2呢？

public boolean search(int[] A, int target) {
        	int low = 0;
            int high = A.length-1;
    		while (low <= high) {
    			int mid = low + (high - low) / 2;
    			if (A[mid] == target) {
    				return true;
    			} else if (A[mid] < A[high]) {
    				if (A[mid] <= target && target <= A[high]) {
    					low = mid+1;
    				} else {
    					high = mid-1;
    				}
    			} else if (A[mid] > A[high]) {
    				if (A[low] <= target && target <= A[mid]) {
    					high = mid-1;
    				} else {
    					low = mid+1;
    				}
    			} else {
    				high--;
    			}
    		}
            return false;
        }

[Link 1]: http://blog.csdn.net/my_jobs/article/details/44114951