题目

给你一个数组 nums ,请你完成两类查询。

其中一类查询要求 更新 数组 nums 下标对应的值
另一类查询要求返回数组 nums 中索引 left 和索引 right 之间( 包含 )的nums元素的 和 ,其中 left <= right
实现 NumArray 类:

NumArray(int[] nums) 用整数数组 nums 初始化对象
void update(int index, int val) 将 nums[index] 的值 更新 为 val
int sumRange(int left, int right) 返回数组 nums 中索引 left 和索引 right 之间( 包含 )的nums元素的 和 (即,nums[left] + nums[left + 1], …, nums[right])

示例 1:

输入:
[“NumArray”, “sumRange”, “update”, “sumRange”]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
输出:
[null, 9, null, 8]

解释:
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9
numArray.update(1, 2); // nums = [1,2,5]
numArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8

提示:

1 <= nums.length <= 3 10^4
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
调用 update 和 sumRange 方法次数不大于 3
10^4

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/range-sum-query-mutable
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

思路

这个题和307. 区域和检索 - 数组可修改 - 图1不同的是,涉及到了数组的更新问题,如果还使用前缀和数组,每更新一处的值就需要影响到前缀和数组的很多值,更新和查询都会达到307. 区域和检索 - 数组可修改 - 图2#card=math&code=O%28n%29&id=xgv5C)时间复杂度。使用「线段树」或者「树状数组」可以将更新和查询操作都降到307. 区域和检索 - 数组可修改 - 图3#card=math&code=O%28logn%29&id=gwg4b)复杂度。真正去学就发现也没有那么难,入门可以参考liweiwei的「文章」和「视频」。

目前我对「线段树」或者「树状数组」只是有一个基本的认识,了解了两者的特点和使用场景。比如「线段树」是一棵「平衡二叉树」(不一定是完全二叉树),「树状数组」是一棵「多叉树」,「线段树」可以用在「区间和查询」(还可以处理区间最值)和「单点更新」问题中,「树状数组」可以处理「前缀和查询」和「单点更新」问题,「树状数组」通过前缀和相减也可以获得区间和。总的来说,树状数组有的功能线段树都有,但反过来就不成立,且线段树代码复杂很多。两者都不适用于区间中有「添加和删除」元素的操作。

需要多应用才能更好地掌握这类进阶的数据结构,这里就不多做介绍了。至于本题,套用「线段树」或者「树状数组」的模板即可解答。

代码

线段树

  1. class NumArray {
  2. private SegmentTree<Integer> segmentTree;
  3. public NumArray(int[] nums) {
  4. Integer[] data = new Integer[nums.length];
  5. for (int i = 0; i < nums.length; i++) {
  6. data[i] = nums[i];
  7. }
  8. segmentTree = new SegmentTree<>(data, (a, b) -> a + b);
  9. }
  10. public void update(int i, int val) {
  11. segmentTree.set(i, val);
  12. }
  13. public int sumRange(int i, int j) {
  14. return segmentTree.query(i, j);
  15. }
  16. }
  17. interface Merge<E> {
  18. E merge(E e1, E e2);
  19. }
  20. class SegmentTree<E> {
  21. private E[] tree;
  22. private E[] data;
  23. private Merge<E> merge;
  24. public SegmentTree(E[] arr, Merge<E> merge) {
  25. this.merge = merge;
  26. data = (E[]) new Object[arr.length];
  27. for (int i = 0; i < arr.length; i++) {
  28. data[i] = arr[i];
  29. }
  30. tree = (E[]) new Object[4 * arr.length];
  31. buildSegmentTree(0, 0, arr.length - 1);
  32. }
  33. private void buildSegmentTree(int treeIndex, int l, int r) {
  34. if (l == r) {
  35. tree[treeIndex] = data[l];
  36. return;
  37. }
  38. int mid = l + (r - l) / 2;
  39. int leftChild = leftChild(treeIndex);
  40. int rightChild = rightChild(treeIndex);
  41. buildSegmentTree(leftChild, l, mid);
  42. buildSegmentTree(rightChild, mid + 1, r);
  43. tree[treeIndex] = merge.merge(tree[leftChild], tree[rightChild]);
  44. }
  45. public E query(int dataL, int dataR) {
  46. if (dataL < 0 || dataL >= data.length || dataR < 0 || dataR >= data.length || dataL > dataR) {
  47. throw new IllegalArgumentException("Index is illegal.");
  48. }
  49. return query(0, 0, data.length - 1, dataL, dataR);
  50. }
  51. private E query(int treeIndex, int l, int r, int dataL, int dataR) {
  52. if (l == dataL && r == dataR) {
  53. return tree[treeIndex];
  54. }
  55. int mid = l + (r - l) / 2;
  56. int leftChildIndex = leftChild(treeIndex);
  57. int rightChildIndex = rightChild(treeIndex);
  58. if (dataR <= mid) {
  59. return query(leftChildIndex, l, mid, dataL, dataR);
  60. }
  61. if (dataL >= mid + 1) {
  62. return query(rightChildIndex, mid + 1, r, dataL, dataR);
  63. }
  64. E leftResult = query(leftChildIndex, l, mid, dataL, mid);
  65. E rightResult = query(rightChildIndex, mid + 1, r, mid + 1, dataR);
  66. return merge.merge(leftResult, rightResult);
  67. }
  68. public void set(int dataIndex, E val) {
  69. if (dataIndex < 0 || dataIndex >= data.length) {
  70. throw new IllegalArgumentException("Index is illegal.");
  71. }
  72. data[dataIndex] = val;
  73. set(0, 0, data.length - 1, dataIndex, val);
  74. }
  75. private void set(int treeIndex, int l, int r, int dataIndex, E val) {
  76. if (l == r) {
  77. tree[treeIndex] = val;
  78. return;
  79. }
  80. int leftTreeIndex = leftChild(treeIndex);
  81. int rightTreeIndex = rightChild(treeIndex);
  82. int mid = l + (r - l) / 2;
  83. if (dataIndex >= mid + 1) {
  84. set(rightTreeIndex, mid + 1, r, dataIndex, val);
  85. }
  86. if (dataIndex <= mid) {
  87. set(leftTreeIndex, l, mid, dataIndex, val);
  88. }
  89. tree[treeIndex] = merge.merge(tree[leftTreeIndex], tree[rightTreeIndex]);
  90. }
  91. public int leftChild(int index) {
  92. return 2 * index + 1;
  93. }
  94. public int rightChild(int index) {
  95. return 2 * index + 2;
  96. }
  97. }
  98. /**
  99. * Your NumArray object will be instantiated and called as such:
  100. * NumArray obj = new NumArray(nums);
  101. * obj.update(index,val);
  102. * int param_2 = obj.sumRange(left,right);
  103. */

树状数组

  1. class NumArray {
  2. FenwickTree fenwickTree;
  3. int[] nums;
  4. public NumArray(int[] nums) {
  5. this.nums = nums;
  6. fenwickTree = new FenwickTree(nums);
  7. }
  8. public void update(int index, int val) {
  9. fenwickTree.update(index + 1, val - nums[index]);
  10. nums[index] = val;
  11. }
  12. public int sumRange(int left, int right) {
  13. return fenwickTree.query(right + 1) - fenwickTree.query(left);
  14. }
  15. }
  16. public class FenwickTree {
  17. private int[] tree;
  18. private int len;
  19. public FenwickTree(int n) {
  20. this.len = n;
  21. tree = new int[n + 1];
  22. }
  23. public FenwickTree(int[] nums) {
  24. this.len = nums.length;
  25. tree = new int[this.len + 1];
  26. for (int i = 1; i <= len; i++) {
  27. update(i, nums[i - 1]);
  28. }
  29. }
  30. public void update(int i, int delta) {
  31. while (i <= len) {
  32. tree[i] += delta;
  33. i += lowbit(i);
  34. }
  35. }
  36. public int query(int i) {
  37. int sum = 0;
  38. while (i > 0) {
  39. sum += tree[i];
  40. i -= lowbit(i);
  41. }
  42. return sum;
  43. }
  44. public static int lowbit(int x) {
  45. return x & (-x);
  46. }
  47. }