LeetCode-Find Minimum in Rotated Sorted Array（找出反转数组中的最小值）

深碍√TFBOYSˉ_ 2022-08-07 03:54 12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`).

Find the minimum element.

You may assume no duplicate exists in the array.

题2：如果允许上题的数组有重复数字呢？

解法：最笨的方法，也是最容易想到的方法就是直接遍历数组，找出最小值，时间复杂度O(n)，但是对于本题明显会超时。此时，比较容易想到的就是[二分查找][Link 1]，因为其有序。但是两者肯定是有不同点的。

第一题的代码如下：

public int findMin(int[] num) {
            int low = 0;
            int high = num.length-1;
            while (low < high-1) {
            	int mid = low + (high-low)/2;
            	if (num[low] < num[high]) {
            		if (num[mid] < num[low]) {
            			low = mid;
            		} else {
            			high = mid;
            		}
            	} else {
            		if (num[mid] < num[high]) {
            			high = mid;
            		} else {
            			low = mid;
            		}
            	}
            }
            return num[low] > num[high] ? num[high] : num[low];
        }

此题一定要区分，这个序列是递增的还是递减的。很多解题方法都没考虑这个问题。显然有了第一题的思路，第二题也就好解了。

public int findMin(int[] num) {
            int low = 0;
            int high = num.length-1;
            while (low < high-1) {
            	int mid = low + (high-low)/2;
            	if (num[low] < num[high]) {
            		if (num[mid] < num[low]) {
            			low = mid;
            		} else if (num[mid] > num[low]){
            			high = mid;
            		} else {
            			low++;
            		}
            	} else {
            		if (num[mid] < num[high]) {
            			high = mid;
            		} else if (num[mid] > num[high]){
            			low = mid;
            		} else {
            			high--;
            		}
            	}
            }
            return num[low] > num[high] ? num[high] : num[low];
        }

[Link 1]: http://blog.csdn.net/my_jobs/article/details/43566661