计算右侧小于当前元素的个数
题目描述
归并排序
代码演示:
ArrayList<Integer> ans = new ArrayList<>();
//记录下标
int[]index;
//记录题目所求的count[i]
int[]count;
public List<Integer> countSmaller(int[] nums) {
int n = nums.length;
index = new int[n];
count = new int[n];
for (int i = 0; i < n;i++){
index[i] = i;
}
mergeSort(nums,0,n - 1);
for (int i = 0; i < n;i++){
ans.add(count[i]);
}
return ans;
}
/**
* 归并排序
* @param nums
* @param l
* @param r
*/
public void mergeSort(int[]nums,int l,int r){
//base case
if (l >= r){
return;
}
int mid = l + (r - l) /2 ;
mergeSort(nums,l,mid);
mergeSort(nums,mid + 1,r);
merge(nums,l,mid,r);
}
/**
* 合并
* @param nums
* @param l
* @param m
* @param r
*/
public void merge(int[]nums,int l,int m,int r){
int[]help = new int[r - l + 1];
int[]tempIndex = new int[r - l + 1];
int i = 0;
//双指针移动 p1 左边部分起始位置,p2 右边部分的起始位置
int p1 = l;
int p2 = m + 1;
//两边开始比较,
while (p1 <= m && p2 <= r){
//help[i++] = nums[p1] <= nums[p2] ? nums[p1++] : nums[p2++];
if (nums[p1] > nums[p2]){
count[index[p1]] += r - p2 + 1;
tempIndex[i] = index[p1];
help[i] = nums[p1];
p1++;
}else{
tempIndex[i] = index[p2];
help[i] = nums[p2];
p2++;
}
i++;
}
//检查哪边没有走完
while (p1 <= m){
help[i] = nums[p1];
tempIndex[i] = index[p1];
p1++;
i++;
}
while (p2 <= r){
help[i] = nums[p2];
tempIndex[i] = index[p2];
p2++;
i++;
}
for (i = 0;i< help.length;i++){
nums[l + i] = help[i];
index[l + i] = tempIndex[i];
}
}
上期经典
leetcode18. 四数之和