由快速排序引出的第k大的数..

//快速排序
void qsort_my(vector<int> & nums, int st, int ed){
if(st > ed)	return;
int key = nums[st];
int i = st, j = ed;
while(i < j){
while(i < j && nums[j] >= key) --j;
nums[i] = nums[j];
while(i < j && nums[i] <= key) ++i;
nums[j] = nums[i];
}
nums[i] = key;
qsort_my(nums, st, i - 1);
qsort_my(nums, j + 1, ed);
}

//单个无序表第K大的数
int KthNum(vector<int> nums, int st, int ed, int k){
int key = nums[st];
int i = st, j = ed;
while(i < j){
while(i < j && nums[j] >= key)	--j;
nums[i] = nums[j];
while(i < j && nums[i] <= key)	++i;
nums[j] = nums[i];
nums[i] = key;
}
nums[i] = key;
if(k == i - st + 1)
return key;
else if(k > i - st + 1)
return KthNum(nums, i + 1, ed, k - i + st - 1);
else
return KthNum(nums, st, i - 1, k);
}

//两个有序表第K大的数
int KthNum2(vector<int> nums1, int st1, int ed1,
vector<int> nums2, int st2, int ed2, int k){
if(st1 > ed1)
return nums2[st2 + k - 1];
else if(st2 > ed2)
return nums1[st1 + k - 1];
else if(k == 1)
return min(nums1[st1], nums2[st2]);
int pt1 = min(k / 2 + st1 - 1, ed1);
int pt2 = min(k / 2 + st2 - 1, ed2);
if(nums1[pt1] >= nums2[pt2])
return KthNum2(nums1, st1, ed1, nums2, pt2 + 1, ed2, k - pt2 + st2 - 1);
else
return KthNum2(nums1, pt1 + 1, ed1, nums2, st2, ed2, k - pt1 + st1 - 1);
}