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303. Range Sum Query - Immutable commit.
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8 changes: 8 additions & 0 deletions
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09-Segment-Tree/303-Range-Sum-Query-Immutable/src/Merger.java
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/** | ||
* @Auther: lss | ||
* @Date: 2019/2/27 20:53 | ||
* @Description: | ||
*/ | ||
public interface Merger<E> { | ||
E merge(E a, E b); | ||
} |
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09-Segment-Tree/303-Range-Sum-Query-Immutable/src/NumArray.java
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/** | ||
* @Auther: lss | ||
* @Date: 2019/3/5 16:58 | ||
* @Description: | ||
*/ | ||
public class NumArray { | ||
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private SegmentTree<Integer> segmentTree; | ||
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public NumArray(int[] nums) { | ||
if (nums.length > 0) { | ||
Integer[] data = new Integer[nums.length]; | ||
for (int i = 0; i < nums.length; i++) | ||
data[i] = nums[i]; | ||
segmentTree = new SegmentTree<>(data, (a, b) -> a + b); | ||
} | ||
} | ||
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public int sumRange(int i, int j) { | ||
if (segmentTree == null) | ||
throw new IllegalArgumentException("Segment Tree is null"); | ||
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return segmentTree.query(i, j); | ||
} | ||
} |
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09-Segment-Tree/303-Range-Sum-Query-Immutable/src/SegmentTree.java
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/** | ||
* @Auther: lss | ||
* @Date: 2019/2/26 20:53 | ||
* @Description: | ||
*/ | ||
public class SegmentTree<E> { | ||
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private E[] tree; | ||
private E[] data; | ||
private Merger<E> merger; | ||
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public SegmentTree(E[] arr, Merger<E> merger) { | ||
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this.merger = merger; | ||
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data = (E[]) new Object[arr.length]; | ||
for (int i = 0; i < arr.length; i++) | ||
data[i] = arr[i]; | ||
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tree = (E[]) new Object[4 * arr.length]; | ||
buildSegmentTree(0, 0, data.length - 1); | ||
} | ||
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// 在treeIndex的位置创建表示区间[l...r]的线段树 | ||
private void buildSegmentTree(int treeIndex, int l, int r) { | ||
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if (l == r) { | ||
tree[treeIndex] = data[l]; | ||
return; | ||
} | ||
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int leftTreeIndex = leftChild(treeIndex); | ||
int rightTreeIndex = rightChild(treeIndex); | ||
// l+r 可能会溢出 所以采用这种 | ||
int mid = l + (r - l) / 2; | ||
buildSegmentTree(leftTreeIndex, l, mid); | ||
buildSegmentTree(rightTreeIndex, mid + 1, r); | ||
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tree[treeIndex] = merger.merge(tree[leftTreeIndex], tree[rightTreeIndex]); | ||
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} | ||
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public int getSize() { | ||
return data.length; | ||
} | ||
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public E get(int index) { | ||
if (index < 0 || index >= data.length) | ||
throw new IllegalArgumentException("Index in illegal"); | ||
return data[index]; | ||
} | ||
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// 返回完全二叉树的数组表示中,一个索引所表示的元素的左孩子节点的索引 | ||
private int leftChild(int index) { | ||
return 2 * index + 1; | ||
} | ||
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// 返回完全二叉树的数组表示中,一个索引所表示的元素的右孩子节点的索引 | ||
private int rightChild(int index) { | ||
return 2 * index + 2; | ||
} | ||
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// 返回区间[queryL,queryR]的值 | ||
public E query(int queryL, int queryR) { | ||
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if (queryL < 0 || queryL >= data.length || | ||
queryR < 0 || queryR >= data.length || queryL > queryR) | ||
throw new IllegalArgumentException("Index is illegal."); | ||
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return query(0, 0, data.length - 1, queryL, queryR); | ||
} | ||
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// 在以treeId为根的线段树中[l...r]的范围里,搜索区间[queryL...queryR]的值 | ||
private E query(int treeIndex, int l, int r, int queryL, int queryR) { | ||
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if (l == queryL && r == queryR) | ||
return tree[treeIndex]; | ||
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int mid = l + (r - l) / 2; | ||
int leftTreeIndex = leftChild(treeIndex); | ||
int rightTreeIndex = rightChild(treeIndex); | ||
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if (queryL >= mid + 1) | ||
return query(rightTreeIndex, mid + 1, r, queryL, queryR); | ||
else if (queryR <= mid) | ||
return query(leftTreeIndex, l, mid, queryL, queryR); | ||
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E leftResult = query(leftTreeIndex, l, mid, queryL, mid); | ||
E rightResult = query(rightTreeIndex, mid + 1, r, mid + 1, queryR); | ||
return merger.merge(leftResult, rightResult); | ||
} | ||
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@Override | ||
public String toString() { | ||
StringBuilder res = new StringBuilder(); | ||
res.append('['); | ||
for (int i = 0; i < tree.length; i++) { | ||
if (tree[i] != null) | ||
res.append(tree[i]); | ||
else | ||
res.append("null"); | ||
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if (i != tree.length - 1) | ||
res.append(", "); | ||
} | ||
res.append(']'); | ||
return res.toString(); | ||
} | ||
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} |