105 lines
2.7 KiB
Java
105 lines
2.7 KiB
Java
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//给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是 nums[i] 右侧小于 num
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//s[i] 的元素的数量。
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//
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//
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// 示例：
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//
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// 输入：nums = [5,2,6,1]
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//输出：[2,1,1,0]
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//解释：
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//5 的右侧有 2 个更小的元素 (2 和 1)
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//2 的右侧仅有 1 个更小的元素 (1)
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//6 的右侧有 1 个更小的元素 (1)
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//1 的右侧有 0 个更小的元素
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//
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//
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//
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//
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// 提示：
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//
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//
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// 0 <= nums.length <= 10^5
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// -10^4 <= nums[i] <= 10^4
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//
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// Related Topics 树状数组 线段树 数组 二分查找 分治 有序集合 归并排序
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// 👍 615 👎 0
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package leetcode.editor.cn;
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import java.util.*;
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//315:计算右侧小于当前元素的个数
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class CountOfSmallerNumbersAfterSelf {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new CountOfSmallerNumbersAfterSelf().new Solution();
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System.out.println(solution.countSmaller(new int[]{1,-3,-2}));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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private int[] c;
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private int[] a;
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public List<Integer> countSmaller(int[] nums) {
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List<Integer> resultList = new ArrayList<Integer>();
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discretization(nums);
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init(nums.length + 5);
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for (int i = nums.length - 1; i >= 0; --i) {
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int id = getId(nums[i]);
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resultList.add(query(id - 1));
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update(id);
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}
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Collections.reverse(resultList);
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return resultList;
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}
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private void init(int length) {
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c = new int[length];
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Arrays.fill(c, 0);
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}
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private int lowBit(int x) {
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return x & (-x);
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}
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private void update(int pos) {
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while (pos < c.length) {
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c[pos] += 1;
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pos += lowBit(pos);
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}
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}
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private int query(int pos) {
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int ret = 0;
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while (pos > 0) {
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ret += c[pos];
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pos -= lowBit(pos);
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}
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return ret;
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}
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private void discretization(int[] nums) {
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Set<Integer> set = new HashSet<Integer>();
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for (int num : nums) {
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set.add(num);
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}
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int size = set.size();
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a = new int[size];
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int index = 0;
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for (int num : set) {
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a[index++] = num;
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}
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Arrays.sort(a);
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}
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private int getId(int x) {
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return Arrays.binarySearch(a, x) + 1;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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