leetcode(153 154) FindMinimuminRotatedSortedArray与FindMinimuminRotatedSortedArray II

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链接:

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/

https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/

两道题的唯一不同点就是有无元素重复

用二分法解.重复情况下仅仅需要多写个else就好了

第一题java代码:

package leetcode;public class FindMinimuminRotatedSortedArray {// https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/// 元素无重复排序// 二分法public static int findMin(int[] nums) {if (nums == null || nums.length == 0)return 0;int l = 0;int r = nums.length - 1;int min = nums[0];while (l < r - 1) {int m = (l + r) / 2;if (nums[l] < nums[m]) {min = Math.min(nums[l], min);l = m + 1;} else{min = Math.min(nums[m], min);r = m - 1;} }min = Math.min(nums[r], min);min = Math.min(nums[l], min);return min;}public static void main(String[] args) {// TODO Auto-generated method stubint[] nums = { 5, 6, 7, 8, 9, 10, 3 };System.out.println(findMin(nums) + "");}}

第二题java代码:

package leetcode;public class FindMinimuminRotatedSortedArrayII {// https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/// 元素有重复排序// 与FindMinimuminRotatedSortedArray不同在于有重复,这是只要多写一个判断条件即可,让左边自加1// 二分法public static int findMin(int[] nums) {if (nums == null || nums.length == 0)return 0;int l = 0;int r = nums.length - 1;int min = nums[0];while (l < r - 1) {int m = (l + r) / 2;if (nums[l] < nums[m]) {min = Math.min(nums[l], min);l = m + 1;} else if (nums[l] > nums[m]) {min = Math.min(nums[m], min);r = m - 1;} else {<span style="white-space:pre"></span>//如果相等,左边自加一次l++;}}min = Math.min(nums[r], min);min = Math.min(nums[l], min);return min;}public static void main(String[] args) {// TODO Auto-generated method stubint[] nums = { 5, 6, 7, 8, 9, 10, 3 };System.out.println(findMin(nums) + "");}}

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