Leetcode_154_Find Minimum in Rotated Sorted Array

本文是在学习中的总结，欢迎转载但请注明出处：http://blog.csdn.net/pistolove/article/details/43416613

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.

思路：

（1）题意为给定一个已经排好序的整形数组，数组的前m(m>0)个元素被移到数组的后面形成新的数组，求新数组中的最小元素。

（2）该题考查的是数组遍历问题。该题同样比较简单，解法一属于“简单暴力”型，直接使用类库中的方法进行排序，取得最小元素（下方算法附有Arrays.sort()中的排序算法实现，感兴趣可以看看，感觉写的好复杂）；解法二属于“一般解法”，通过从前、从后遍历求得最小值，从后往前的效率要高于从前往后；解法三应该是属于"高效性"，不过这里尚未给出实现，猜测的思路是通过类似“二分查找”的方式来寻求最小值，感兴趣的可以研究下。

（3）希望本文对你有所帮助。

算法代码实现如下：

/** * @author liqq * 解法一：简单暴力 */public int findMin(int[] num) {if (num == null || num.length == 0)return 0;Arrays.sort(num);return num[0];}

/** *  * @author liqq 附加Arrays.sort() 源码 以供参考 */public int findMin(int[] num) {if (num == null || num.length == 0)return 0;sort(num, 0, num.length);return num[0];}private static void sort(int x[], int off, int len) {// Insertion sort on smallest arraysif (len < 7) {for (int i = off; i < len + off; i++)for (int j = i; j > off && x[j - 1] > x[j]; j--)swap(x, j, j - 1);return;}// Choose a partition element, vint m = off + (len >> 1); // Small arrays, middle elementif (len > 7) {int l = off;int n = off + len - 1;if (len > 40) { // Big arrays, pseudomedian of 9int s = len / 8;l = med3(x, l, l + s, l + 2 * s);m = med3(x, m - s, m, m + s);n = med3(x, n - 2 * s, n - s, n);}m = med3(x, l, m, n); // Mid-size, med of 3}int v = x[m];// Establish Invariant: v* (<v)* (>v)* v*int a = off, b = a, c = off + len - 1, d = c;while (true) {while (b <= c && x[b] <= v) {if (x[b] == v)swap(x, a++, b);b++;}while (c >= b && x[c] >= v) {if (x[c] == v)swap(x, c, d--);c--;}if (b > c)break;swap(x, b++, c--);}// Swap partition elements back to middleint s, n = off + len;s = Math.min(a - off, b - a);vecswap(x, off, b - s, s);s = Math.min(d - c, n - d - 1);vecswap(x, b, n - s, s);// Recursively sort non-partition-elementsif ((s = b - a) > 1)sort(x, off, s);if ((s = d - c) > 1)sort(x, n - s, s);}private static void swap(int x[], int a, int b) {int t = x[a];x[a] = x[b];x[b] = t;}private static int med3(int x[], int a, int b, int c) {return (x[a] < x[b] ? (x[b] < x[c] ? b : x[a] < x[c] ? c : a): (x[b] > x[c] ? b : x[a] > x[c] ? c : a));}private static void vecswap(int x[], int a, int b, int n) {for (int i = 0; i < n; i++, a++, b++)swap(x, a, b);}

/** * @author liqq  * 解法二 分别从前、从后遍历 从后遍历效率稍微高一些 */public int findMinFromHead(int[] num) {if (num == null || num.length == 0)return 0;int len = num.length;int leftHalfMin = num[0];for (int i = 1; i < len; i++) {if (leftHalfMin >= num[i]) {leftHalfMin = num[i];}}return leftHalfMin;}public int findMinFromEnd(int[] num) {if (num == null || num.length == 0)return 0;if (num.length == 1)return num[0];int len = num.length;int min = num[len - 1];for (int i = len - 2; i >= 0; i--) {if (min <= num[i]) {return min;} else {min = num[i];if (i == 0) {return min;}}}return -1;}