第四章 快速排序 分而治之(divide an conquer)

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for i in range(1,len(array)):     if array[i] <= flag:         less.append(array[i])     else:         greater.append(array[i])       return quicksort(less) + [flag] + quicksort(greater) for i in range(1,len(array)):     if array[i] <= flag:         less.append(array[i])     else:         greater.append(array[i])       return quicksort(less) + [flag] + quicksort(greater)

## 分而治之 (devide and conquer) 1. 找出基线条件,这种条件必须尽可能简单. 2. 不断将问题缩小规模,知道符合基线条件.