Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

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最新推荐文章于 2022-05-26 16:47:57 发布

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原文链接：http://www.cnblogs.com/codingEskimo/p/6805093.html

本文介绍了一种在已旋转的排序数组中查找最小元素的方法。通过使用二分搜索算法，该方法能在O(log N)的时间复杂度下找到数组中的最小值。文章详细解释了如何在遇到重复元素时优化搜索过程。

The worst situation O(N).

Actually we can either just loop through, or we can compare num[mid] with the num[end], if they are the same, that means it's fine to remove end, the smallest element won't be removed

public class Solution {
    /**
     * @param num: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] num) {
        // write your code here
        if (num == null || num.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = num.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            // we want to keep the lower part, so use endVal to compare
            if (num[mid] == num[end]) {
                end--;
            } else if (num[mid] < num[end]) {
                end = mid;
            } else {
                start = mid;
            }
        }
        
        if (num[start] <= num[end]) {
            return num[start];
        }
        
        if (num[end] < num[start]) {
            return num[end];
        }
        return -1;
    }
}