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.
解法
二分查找,总是和最后一个数去比较num[end],如果大,则在数组的左半区域,需要往右半部分走;如果小,则在数组的右半区域,需要往左走。
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0;
int end = nums.length - 1;
int mid;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (nums[mid] < nums[end]) {
end = mid;
} else {
start = mid;
}
}
return nums[start] < nums[end] ? nums[start] : nums[end];
}
}
本文介绍了一种在未知旋转点的升序数组中寻找最小元素的方法。通过二分查找算法,每次比较中间元素与最后一个元素来确定搜索区间,最终找到最小值。
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