Wednesday, November 12, 2014

[LeeCode] Find Minimum in Rotated Sorted Array I, II

Find Minimum in Rotated Sorted Array I

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.

Find Minimum in Rotated Sorted Array II

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.
The array may contain duplicates.

思路:Find Minimum in Rotated Sorted Array I
Search in Rotated Sorted Array I这题换汤不换药。同样可以根据A[mid]和A[end]来判断右半数组是否sorted:
原数组:0 1 2 4 5 6 7
情况1:  6 7 0 1 2 4 5   
情况2:  2 4 5 6 7 0 1  
(1) A[mid] < A[end]:A[mid : end] sorted => min不在A[mid+1 : end]中
搜索A[start : mid]
(2) A[mid] > A[end]:A[start : mid] sorted且又因为该情况下A[end]<A[start] => min不在A[start : mid]中
搜索A[mid+1 : end]
(3) base case:
a. start =  end,必然A[start]为min,为搜寻结束条件。
b. start + 1 = end,此时A[mid] =  A[start],而min = min(A[mid], A[end])。而这个条件可以合并到(1)和(2)中。

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class Solution {
public:
    int findMin(vector<int> &num) {
        int start = 0, end = num.size()-1;
        while(start<end) {
            int mid = start+(end-start)/2;
            if(num[mid]<num[end]) 
                end = mid;
            else 
                start = mid+1;
        }
        return num[start];
    }
};


思路:Find Minimum in Rotated Sorted Array II

和Search in Rotated Sorted Array II这题换汤不换药。同样当A[mid] = A[end]时,无法判断min究竟在左边还是右边。

3 1 2 3 3 3 3 
3 3 3 3 1 2 3

但可以肯定的是可以排除A[end]:因为即使min = A[end],由于A[end] = A[mid],排除A[end]并没有让min丢失。所以增加的条件是:

A[mid] = A[end]:搜索A[start : end-1]


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class Solution {
public:
    int findMin(vector<int> &num) {
        int start = 0, end = num.size()-1;
        while(start<end) {
            int mid = start + (end-start)/2;
            if(num[mid]<num[end]) 
                end = mid;
            else if(num[mid]>num[end])
                start = mid+1;
            else
                end--;
        }
        return num[start];
    }
};

2 comments:

  1. 你好,有个更一般的问题。在做binary sesarch的时候,循环条件怎么设定?start < end 还是start <= end,
    还有怎么缩放边界,end = mid 还是end = mid -1 .这个我一直无法把握,请问有什么技巧?谢谢

    ReplyDelete