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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?[4, 5, 1, 2, 3]
-> [1,
2, 3, 4, 5]
In-place, O(1) extra space and O(n) time.
What is rotated array?
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