leetcode解題之153&154. Find Minimum in Rotated Sorted Array版(在旋轉的陣列中查詢最小數字)
153.Find Minimum in Rotated Sorted Array
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
).
Find the minimum element.
You may assume no duplicate exists in the array.
給定一個數組,陣列可能被旋轉了,但是在旋轉之前是遞增有序的,找出該陣列的的最小值。(陣列沒有重複值
複雜度等價於一個二分查詢,是O(logn),空間上只有四個變數維護二分和結果,所以是O(1)。
採用二分查詢的方法進行查詢。陣列的第一個元素要大於最後一個元素,因為陣列是遞增的,如果不小於則第一個元素就是最小的元素。如果不小於,初始化high=0,high=len-1,則取中間數mid,如果中間數大於high指向的元素,則說明小數存在後半段,則將high指向mid,否則將high指向mid,繼續進行尋找。
只需要判斷是否為遞增序列或者是否只有一個元素特殊情況即可,新增一句,if (nums[low] <= nums[high]) return nums[low];繼續二分查詢。等號判斷只有一個元素情況,小於判斷遞增序列。注意:如果
154. Find Minimum in Rotated Sorted Array II
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 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
).
Find the minimum element.
The array may contain duplicates.
給定一個數組,陣列可能被旋轉了,但是在旋轉之前是遞增有序的,找出該陣列的的最小值。(陣列有重複值)
兩道題解法一樣,有重複值只要判斷 low high mid三者指向的元素是否相等,相等就退化為O(N)複雜度。
public int search(int[] nums, int target) {
if (nums == null || nums.length < target)
return 0;
int low = 0;
int res = 0;
int high = nums.length - 1;
if (nums[low] <= nums[high])
res = Arrays.binarySearch(nums, target);
else {
int m = findMin(nums);
System.out.println(m);
res = Arrays.binarySearch(nums, 0, m, target);
if (res < 0)
res = Arrays.binarySearch(nums, m, high + 1, target);
}
if (res < 0)
return -1;
else
return res;
}
public int findMin(int[] nums) {
// ask interviewer which value should return:
// Integer.MIN_VALUE or throw a Exception.
if (nums == null || nums.length == 0)
return -1;
int low = 0;
int mid = 0;
int high = nums.length - 1;
while (nums[low] >= nums[high]) {
// 表示找到
if (high - low == 1) {
mid = high;
break;
}
mid = (low + high) / 2;
// 如果三個數字相等需要遍歷整個數字查詢最小值
if (nums[mid] == nums[low]
&& nums[mid] == nums[high])
return MinOrder(nums, low, high);
// 處於遞增子序列中,最小值在右側
if (nums[mid] >= nums[low])
low = mid;
else
// 處於遞減子序列中,最小值在左側
high = mid;
}
return nums[mid];
}
// 遍歷整個陣列求最小值
private int MinOrder(int[] nums, int low, int high) {
int min = nums[low];
for (int i = low + 1; i <= high; i++)
min = Math.min(min, nums[i]);
return min;
}
網上參考:http://www.programcreek.com/2014/03/leetcode-find-minimum-in-rotated-sorted-array-ii-java/public class Solution {
public int findMin(int[] num) {
return findMin(num, 0, num.length-1);
}
public int findMin(int[] num, int left, int right){
if(right==left){
return num[left];
}
if(right == left +1){
return Math.min(num[left], num[right]);
}
// 3 3 1 3 3 3
int middle = (right-left)/2 + left;
// already sorted
if(num[right] > num[left]){
return num[left];
//right shift one
}else if(num[right] == num[left]){
return findMin(num, left+1, right);
//go right
}else if(num[middle] >= num[left]){
return findMin(num, middle, right);
//go left
}else{
return findMin(num, left, middle);
}
}