110.Find Minimum in Rotated Sorted Array
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Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7
might become 4 5 6 7 0 1 2
).
Find the minimum element.
You may assume no duplicate exists in the array.
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分析:即找经有序数组旋转之后得到的数组的最小元素。
给定的序列是部分有序,当输入的序列为空[]时返回0.
分别用两个指针分别指向序列的第一个元素和最后一个元素。
进入下面四个步骤的循环的条件是nums[index1]>=nums[index2].
Step1:计算出中间的元素下标为mid = index1+(index2-index1)/2;
Step2:如果nums[index1]==nums[index2] && nums[index1]==nums[mid],即序列的第一个元素、中间元素和最后一个元素相等,则按照循环查找返回数组中的最小元素;
Step3:如果nums[index1]<=nums[mid],则最小元素在序列的后半部分,令index1 = mid,查找的范围缩短为后半部分,转向Step1;否则转向Step4.
Step4: 则最小元素在序列的前半部分,令index2 = mid,查找的范围缩短为前半部分,转向Step1。
/** * 求旋转数组中的最小元素。 */ public int findMin(int [] array) { if(array.length==0){ return 0; } int index1 = 0;//指向序列中的第一个元素 int mid = index1;//指向序列中的中间元素 int index2 = array.length-1;//指向序列中的最后一个元素 while(array[index1]>=array[index2]){ if (index2-index1 == 1) { mid = index2; break;}mid = index1+(index2-index1)/2;/* * 如果nums[index1]==nums[index2] && nums[index1]==nums[mid], * 即序列的第一个元素、中间元素和最后一个元素相等,则按照循环查找返回数组中的最小元素 */if(array[index1] == array[mid] && array[index2] == array[mid]){return FindMinInOrder(array,index1,index2);}/* * 如果nums[index1]<=nums[mid],则最小元素在序列的后半部分,令index1 = mid,查找的范围缩短为后半部分 */if(array[index1] <= array[mid]){index1 = mid;}/* * 否则最小元素在序列的前半部分,令index2 = mid,查找的范围缩短为前半部分,转向Step1 */else{index2 = mid;} }return array[mid]; } /** * 顺序查找序列中的最小元素 . * 101111、111101等情况 */ private int FindMinInOrder(int [] array,int index1,int index2){ int re = array[index1]; for(int i=index1;i<=index2;i++){ if(re>array[i]){ re = array[i]; } } return re; }
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