algorithm - 返回枢轴位置的就地分区

Question

由于正常分区返回一个索引 j ，使得索引 i <= j 的每个元素都小于选择的枢轴，并且索引 m > j 的每个元素都大于枢轴，因此不能保证 j 是枢轴。是否有可能创建另一个准确返回新枢轴位置的就地分区算法？最初，我坚持将选择的枢轴移动到最后一个位置，但这并没有导致最佳解决方案。

Answer 1

关键是要使枢轴不被交换，除非在开始时将其存储在其中一个边界处，最后将其交换到正确的位置时除外。

int partition(int *A, int low, int high) {

    if (high <= low) return low; // in that case, partition shouldn't be called, but cater for it

    int pivot_position = low;  // or high, or low + (high-low)/2, or whatever
    int pivot = A[pivot_position];

    A[pivot_position] = A[high];
    A[high] = pivot;    // these two lines are unnecessary if pivot_position == high
    int i = low, j = high-1;

    while(i < j) {

        while(i < j && A[i] <= pivot)
            ++i;   // i == j or A[i] > pivot, and A[k] <=pivot for low <= k < i

        while(i < j && A[j] > pivot) 
            --j;   // i == j or A[j] <= pivot, and A[m] > pivot for j < m < high

        if (i < j) {
            int temp = A[j];
            A[j] = A[i];
            A[i] = temp;
        }
    }
    if (A[i] <= pivot) 
        ++i;

    // now A[k] <= pivot for low <= k < i and A[m] > pivot for i <= m < high
    A[high] = A[i];
    A[i] = pivot;
    return i;
}