leetcode_middle_86_81. Search in Rotated Sorted Array II
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题意:
将一个升序数组,从中切断,将切断点后面的序列拼接到前面。注:切断点也可能在最后一个数字之后,即数组保持升序不变。
注意:数组中可能有重复数字。
分析:
这个数组长什么样呢?
* *
*
*
*
*
* *
*
可能是长这样的,甚至长这样。* * * * * * * *
我们怎么能确定mid落在哪个区间呢?通过和nums[ low ] 或者nums[ high ]比较是可以得到的。但是原来的大于等于或者小于等于就不一定得到了。必须要大于或者小于才能充分证明可以得到。
至于当等于的时候, 比如 nums[ low ] == nums[ mid ]
说明可能是这样:
*
* * * *
low mid
我们可以low++,来打破这种相等,从而可以判断区间,当然如果一直无法打破,那么就是全等的极端情况,只能依赖low++的顺序搜索, 最坏情况时间复杂度logn变n了。
public class Solution { public boolean search(int[] nums, int target) { int low = 0; int high = nums.length-1; while(low<=high){ int mid = low + (high - low)/2; if(nums[mid] == target) return true; if(nums[mid] > nums[low]){ //则一定在左上区间 if(target>=nums[low] && target<nums[mid]){ high = mid - 1; }else{ low = mid + 1; } }else if(nums[mid] < nums[low]){ //则一定在右下区间 if(target>nums[mid] && target<=nums[high]){ low = mid + 1; }else{ high = mid - 1; } }else{ low++; } } return false; }}
public class Solution { public boolean search(int[] nums, int target) { int start = 0, end = nums.length - 1; while(start <= end){ int mid = start + (end - start)/2; if(nums[mid] == target) return true; if(nums[start] < nums[mid]){ if(target < nums[start] || target > nums[mid]){ start = mid + 1; }else{ end = mid - 1; } }else if(nums[start] > nums[mid]){ if(target < nums[mid] || target > nums[end]){ end = mid -1; }else{ start = mid + 1; } } else{ start++; } } return false;}}
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