#464 Sort Integers II

题目描述：

Given an integer array, sort it in ascending order. Use quick sort, merge sort, heap sort or any O(nlogn) algorithm.

Have you met this question in a real interview? 

Yes

Example

Given [3, 2, 1, 4, 5], return [1, 2, 3, 4, 5].

题目思路：

今天喜欢merge sort，就用merge sort好了。需要注意的是，在merge两个sorted array时，需要开辟一个新的array用来存原来的A中start~end的值，否则直接在A上做in-place是会wrongly overwrite的。这个新的array必须是长度为end-start+1的，之前我偷懒直接开了一个A的copy，结果就不能pass lintcode的空间复杂度了。

Mycode（AC = 501ms）：

class Solution {
public:
    /**
     * @param A an integer array
     * @return void
     */
    void sortIntegers2(vector<int>& A) {
        // Write your code here
        mergeSort(A, 0, A.size() - 1);
    }
    
    void mergeSort(vector<int>& A, int start, int end) {
        if (start >= end) return;
        
        // sort the left half and right half in A
        mergeSort(A, start, (start + end) / 2);
        mergeSort(A, (start + end) / 2 + 1, end);
        
        // merge the two sorted arrays
        merge(A, start, (start + end) / 2, end);
    }
    
    void merge(vector<int>& A, int start, int mid, int end) {
        // use a tmp array to store A:start ~ end
        vector<int> tmp(end - start + 1, 0);
        int A_idx = start;
        for (int i = 0; i < tmp.size(); i++) {
            tmp[i] = A[A_idx++];
        }
        
        int l = 0, r = mid + 1 - start, idx = start;
        // merge the sorted arrays
        while (l <= mid - start && r < tmp.size()) {
            if (tmp[l] <= tmp[r]) {
                A[idx++] = tmp[l++];
            }
            else {
                A[idx++] = tmp[r++];
            }
        }
        
        while (l <= mid - start) A[idx++] = tmp[l++];
        while (r < tmp.size()) A[idx++] = tmp[r++];
    }
};