Count Of Smaller Number
Count Of Smaller Number
题目地址
http://www.lintcode.com/problem/count-of-smaller-number/description
题目描述
Give you an integer array (index from 0 to n-1, where n is the size of this array, value from 0 to 10000) and an query list. For each query, give you an integer, return the number of element in the array that are smaller than the given integer.
Example
Example 1:
Input: array =[1,2,7,8,5] queries =[1,8,5]
Output:[0,4,2]
Example 2:
Input: array =[3,4,5,8] queries =[2,4]
Output:[0,1]
Challenge
Could you use three ways to do it.
Just loop
Sort and binary search
Build Segment Tree and Search.
代码
public class Solution {
class SegmentTreeNode {
public int start, end;
public int count;
public SegmentTreeNode left, right;
public SegmentTreeNode(int start, int end, int count) {
this.start = start;
this.end = end;
this.count = count;
this.left = this.right = null;
}
}
SegmentTreeNode root;
public SegmentTreeNode build(int start, int end) {
if (start > end) return null;
SegmentTreeNode root = new SegmentTreeNode(start, end, 0);
if (start != end) {
int mid = (start + end) / 2;
root.left = build(start, mid);
root.right = build(mid + 1, end);
} else {
root.count = 0;
}
return root;
}
public int querySegmentTree(SegmentTreeNode root, int start, int end) {
if (start == root.start && root.end == end) return root.count;
int mid = (root.start + root.end) / 2;
int leftCount = 0, rightCount = 0;
if (start <= mid) {
if (mid < end) {
leftCount = querySegmentTree(root.left, start, mid);
} else {
leftCount = querySegmentTree(root.left, start, end);
}
}
if (mid < end) {
if (start <= mid) {
rightCount = querySegmentTree(root.right, mid + 1, end);
} else {
rightCount = querySegemntTree(root.right, start, end);
}
}
return leftCount + rightCount;
}
public void modifySegmentTree(SegmentTreeNode root, int index, int value) {
if (root.start == index && root.end == index) {
root.count += value;
return;
}
int mid = (root.start + root.end) / 2;
if (root.start <= index && index <= mid) {
modifySegemntTree(root.left, index. vlaue);
}
if (mid < index && index <= root.end) {
modifySegmentTree(root.right, index, value);
}
root.count = root.left.count + root.right.count;
}
public ArrayList<Integer> countOfSmallerNumber(int[] A, int[] queries) {
root = build(0, 10000);
ArrayList<Integer> ans = new ArrayList<Integer>();
int res;
for (int i = 0; i < A.length; i++) {
modifySegmentTree(root, A[i], 1);
}
for (int i = 0; i < queries.length; i++) {
res = 0;
if (queries[i] > 0) {
res = querySegmentTree(root, 0, queries[i] - 1);
}
ans.add(res);
}
return ans;
}
}
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