Find Minimum in Rotated Sorted Array II
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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.
思路: 用二分法来进行判断, 注意 0, 1, 2, 这种纯升序的情况。 每次一共有三种情况, num[M] > num[R] , 在右半部分。 num[M] <= num[R] && num[M] < num[L] , 在左半部分。 还有一种就是 L == M == R , 这种情况下, L 向右移动一位。 最后返回num[L]
易错点: 1. L < R && num[L] >= num[R]
2. 在左半部分时候 , R = M;
public class Solution { public int findMin(int[] num) { int L = 0, R = num.length - 1; while (L < R && num[L] >= num[R]) { int M = (L + R) / 2; if (num[M] > num[R]) { L = M + 1; } else if (num[M] < num[L]) { R = M;//----- } else { // A[L] == A[M] == A[R] L = L + 1; } } return num[L]; }}
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- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array II
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