本文主要是介绍find peak or drop，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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找打peak or drop，原来用的是递归，这次改为迭代
public static int helper1(int[] nums) {if (nums.length == 0)return -1;int len = nums.length;if (Math.abs(nums[len - 1] - nums[0]) == len - 1)return -1;// There is no such a valley or peakint start = 0, end = nums.length - 1;while (start < end) {int mid = start + (end - start) / 2;if ((nums[mid - 1] - nums[mid]) * (nums[mid + 1] - nums[mid]) > 0)return nums[mid];int diffOfIndex = mid - start;int diffOfValue = Math.abs(nums[mid] - nums[start]);//因为如果不相差，则下标和值是对应的，则没有发生波峰或波谷if (diffOfIndex == diffOfValue)start = mid;elseend = mid;}return -1;}public static void main(String[] args) {int[] nums1 = { 1, 2, 3, 4, 3, 2 };int[] nums2 = {};int[] nums3 = { 1, 2, 3, 4, 5, 6, 7, 8, 7 };int[] nums4 = { 1, 2, 3, 4, 5 };int[] nums5 = { 5, 4, 3, 2, 1, 2, 3 };int[] nums6 = { 9, 8, 7, 6, 7 };System.out.println(find(nums1));System.out.println(find(nums2));System.out.println(find(nums3));System.out.println(find(nums4));System.out.println(find(nums5));System.out.println(find(nums6));System.out.println(helper1(nums1));System.out.println(helper1(nums2));System.out.println(helper1(nums3));System.out.println(helper1(nums4));System.out.println(helper1(nums5));System.out.println(helper1(nums6));}这篇关于find peak or drop的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
