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lintcode Recover Rotated Sorted Array

时间:2015-08-03 19:14:31      阅读:160      评论:0      收藏:0      [点我收藏+]

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Given a rotated sorted array, recover it to sorted array in-place.

Have you met this question in a real interview? 
Yes
Example

[4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5]

Challenge

In-place, O(1) extra space and O(n) time.

Clarification

What is rotated array?

  • For example, the orginal array is [1,2,3,4], The rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]

zh找到轴点,分别交换三次
class Solution {
public:
    void recoverRotatedSortedArray(vector<int> &nums) {
        // write your code here
        int len = nums.size();
        if (0 == len) {
            return ;
        }
        int t = 0;
        for (int i = 0; i < len; ++i) {
            if (nums[i] < nums[t] ) {
                t = i;
            }
        }
        reverse(nums,0,t-1);
        reverse(nums,t,len-1);
        reverse(nums,0,len-1);
    }
    void reverse(vector<int> &nums,int lo,int hi) {
        while (lo<hi) {
            swap(nums[lo],nums[hi]);
            ++lo;
            --hi;
        }
    }
};


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lintcode Recover Rotated Sorted Array

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原文地址:http://blog.csdn.net/susser43/article/details/47259409

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