Merge Sort

Sort 的时间复杂度：o(n),  空间复杂度 o(logn)。

Merge 的时间复杂度: o(nlogn), 空间复杂度 o(n）。

总时间复杂度：o(nlogn)  总空间复杂度：o(n）。

public class Solution {
  public int[] mergeSort(int[] array) {
    // Write your solution here
    if (array == null) {
      return array;//check null array first.
  }
    //allocate helper array to help merge sort.
    //so that we guanrantee no more than 0(n) space is used.
    //The space complexity is O(n) in this case.
    int[] helper = new int[array.length];//把helper放前面，会节省运行的空间。
    mergeSort(array, helper , 0, array.length - 1);
    return array;
  }
  private void mergeSort(int[] array, int[] helper, int left, int right) {
    if (left >= right) {
      return;
    }
    int mid = left + (right - left)/2;//find the mid
    mergeSort(array, helper, left, mid);
    mergeSort(array, helper, mid+1, right);
    merge(array, helper, left, mid, right);
  }
  private void merge(int[] array, int[] helper, int left, int mid, int right) {
    //copy the content to helper array and we will merge from the 
    //helper array.
    for (int i = left; i <= right; i++) {
      helper[i] = array[i];
}
    int leftIndex = left;
    int rightIndex = mid+1;
    while (leftIndex <= mid && rightIndex <= right) {
      if (helper[leftIndex] <= helper[rightIndex]) {
        array[left++] = helper[leftIndex++];
      } else {
        array[left++] = helper[rightIndex++];
      }
          }
     //if we still have some elements at left side, we need to copy them
     while (leftIndex <= mid) {
       array[left++] = helper[leftIndex++];
       //if there are some elements at right side,we do not need to copy them
       //because they are already in their position.
       
     }
    
}
  
}