【LeetCode】【Solution】Find Minimum in Rotated Sorted Array I & II
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今天第一次看到LeetCode OJ右上角 Solution 上那个小锁被打开了,虽然是看了别人的代码才做到的,但还是很激动。这个小锁里面可是经典的解法啊,如同解开了宝藏的密码。以后当多多努力!
【题目】
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.
I: You may assume no duplicate exists in the array.
II: The array may contain duplicates.【I】
public class Solution { public int findMin(int[] num) { int L = 0, R = num.length-1; while (num[L] > num[R]) { int M = (L + R) / 2; if (num[M] > num[R]) { L = M + 1; } else { R = M; // be careful, not mid-1 } } return num[L]; }}
【II】
public int findMin(int[] A) { int L = 0, R = A.length - 1; while (L < R && A[L] >= A[R]) { int M = (L + R) / 2; if (A[M] > A[R]) { L = M + 1; } else if (A[M] < A[L]) { R = M; } else { // A[L] == A[M] == A[R] L = L + 1; } } return A[L];}
【总结】
上面的解法是LeetCode提供的“标准”答案,当然也非常简洁。
主要考虑中间位置M和数组rotate的位置之间的关系,II还要考虑连续相等的情况。
如果只看【II】的第三种情况应该是 A[L] <= A[M] <= A[R],而非 A[L] == A[M] == A[R],但是while的条件限制了 A[L] >= A[R],所以只有可能是A[L] == A[R],也即A[L] == A[M] == A[R]。
下图可帮助理解:
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