标签：leetcode rotate array binary search
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.
这题其实就是找翻转后右侧部分的最左端点，典型的二分搜索变体，大致思路分成两种情况讨论：
1）如果搜索范围在翻转部分，稍微复杂一点：
1.1）如果中间点在翻转后的左侧，则应该往右寻找，因为如果没有冗余元素，最小值不可能出现在左侧；
1.2）如果中间点在翻转后的右侧，则应该往左寻找。不过这时应该缓存下当前中间点对应的元素，因为再往左的话，可能就到了翻转后的左侧部分，就可能错过全局最小值了。
2）如果搜索范围在排序的部分，则当前的最小值肯定是在最左端。不过这个时候最左端的值不一定是全局最小，需要跟缓存的最小值在进行一次比较。
public int findMin(int[] num) {
int low = 0, high = num.length - 1;
int ret = Integer.MAX_VALUE;
while (low <= high) {
int mid = low + (high - low) / 2;
// On the sorted part.
if (num[low] <= num[high]) {
return Math.min(num[low], ret);
} else { // On the rotated part.
// num[mid] is on the left rotated part.
if (num[mid] >= num[low]) {
low = mid + 1;
} else { // num[mid] is on the right rotated part.
ret = num[mid];
high = mid - 1;
}
}
}
return ret;
}
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
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.
The array may contain duplicates.
如果存在冗余元素，那么则必须把num[low]和num[high]相等的情况单独讨论了。这种情况下，只能老老实实的线性搜索了。如果使用二分查找，这个时候可能会出问题：如果输入数组的最左和最右元素相等，而且都存在一片连续相等的区域，那么中间点可能也会在最左或者最右的连续相等区域中。这个时候应该是无法判断再往哪个方向搜索的。例如[3, 3, 3, 3, 3, 3, 4, 2, 3, 3]和[3, 3, 4, 2, 3, 3, 3, 3, 3, 3]，取中间点后根本无法判断继续往哪个方向搜索。
public int findMin(int[] num) {
int low = 0, high = num.length - 1;
int ret = Integer.MAX_VALUE;
while (low <= high) {
// Linear scan to find a non-flat part.
if (num[low] == num[high]) {
ret = Math.min(num[low], ret);
low++;
continue;
}
int mid = low + (high - low) / 2;
// On the sorted part.
if (num[low] < num[high]) {
return Math.min(num[low], ret);
} else { // On the rotated part.
// num[mid] is on the left rotated part.
if (num[mid] >= num[low]) {
low = mid + 1;
} else { // num[mid] is on the right rotated part.
ret = num[mid];
high = mid - 1;
}
}
}
return ret;
}[LeetCode] Find Minimum in Rotated Sorted Array
标签：leetcode rotate array binary search
原文地址：http://blog.csdn.net/whuwangyi/article/details/40423895
