public class Solution { public boolean search(int[] nums, int target) { if (nums == null || nums.length == 0) return false; int left = 0, right = nums.length - 1; int mid = 0; while (left <= right) { mid = left + (right - left) / 2; if (nums[mid] == target) return true; else if(nums[left] < nums[mid]) { if (nums[left] == target) return true; else if (nums[left] < target && target < nums[mid]) right = mid - 1; else left = mid + 1; } else if (nums[left] == nums[mid]) { left++; } else { if (nums[right] == target) return true; else if (nums[mid] < target && target < nums[right]) left = mid + 1; else right = mid - 1; } } return false; } }
Leetcode - Search in Rotated Sorted Array II
[分析] 同Search in Rotated Sorted Array 的差别在于数组元素可能重复,遇到重复无法推断target所属范围时就一步步移动左边界直到左边界和中间元素不同或者相遇为止。联系Find Minimum in Rotated Sorted Array II进行理解。
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