线段树 04 Leetcode中的303号问题

303. 区域和检索 - 数组不可变

• 线段树解法

class NumArray {
    
    private interface Merger {
        E merge(E a, E b);
    }

    private class SegmentTree {

        private E[] tree;
        private E[] data;
        private Merger merger;

        public SegmentTree(E[] arr, Merger merger){

            this.merger = merger;

            data = (E[])new Object[arr.length];
            for(int i = 0 ; i < arr.length ; i ++)
                data[i] = arr[i];

            tree = (E[])new Object[4 * arr.length];
            buildSegmentTree(0, 0, arr.length - 1);
        }

        // 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){

            if(l == r){
                tree[treeIndex] = data[l];
                return;
            }

            int leftTreeIndex = leftChild(treeIndex);
            int rightTreeIndex = rightChild(treeIndex);

            // int mid = (l + r) / 2;
            int mid = l + (r - l) / 2;
            buildSegmentTree(leftTreeIndex, l, mid);
            buildSegmentTree(rightTreeIndex, mid + 1, r);

            tree[treeIndex] = merger.merge(tree[leftTreeIndex], tree[rightTreeIndex]);
        }

        public int getSize(){
            return data.length;
        }

        public E get(int index){
            if(index < 0 || index >= data.length)
                throw new IllegalArgumentException("Index is illegal.");
            return data[index];
        }

        // 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;
        }

        // 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;
        }

        // 返回区间[queryL, queryR]的值
        public E query(int queryL, int queryR){

            if(queryL < 0 || queryL >= data.length ||
                    queryR < 0 || queryR >= data.length || queryL > queryR)
                throw new IllegalArgumentException("Index is illegal.");

            return query(0, 0, data.length - 1, queryL, queryR);
        }

        // 在以treeIndex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){

            if(l == queryL && r == queryR)
                return tree[treeIndex];

            int mid = l + (r - l) / 2;
            // treeIndex的节点分为[l...mid]和[mid+1...r]两部分

            int leftTreeIndex = leftChild(treeIndex);
            int rightTreeIndex = rightChild(treeIndex);
            if(queryL >= mid + 1)
                return query(rightTreeIndex, mid + 1, r, queryL, queryR);
            else if(queryR <= mid)
                return query(leftTreeIndex, l, mid, queryL, queryR);

            E leftResult = query(leftTreeIndex, l, mid, queryL, mid);
            E rightResult = query(rightTreeIndex, mid + 1, r, mid + 1, queryR);
            return merger.merge(leftResult, rightResult);
        }

        @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");

                if(i != tree.length - 1)
                    res.append(", ");
            }
            res.append(']');
            return res.toString();
        }
    }

    private SegmentTree segmentTree;
    
    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);
        }
    }
    
    public int sumRange(int i, int j) {
        return segmentTree.query(i, j);
    }
}

• 数组解法

  public class NumArray {

    private int[] sum; // sum[i]存储前i个元素和, sum[0] = 0
                       // 即sum[i]存储nums[0...i-1]的和
                       // sum(i, j) = sum[j + 1] - sum[i]
    public NumArray(int[] nums) {

        sum = new int[nums.length + 1];
        sum[0] = 0;
        for(int i = 1 ; i < sum.length ; i ++)
            sum[i] = sum[i - 1] + nums[i - 1];
    }

    public int sumRange(int i, int j) {
        return sum[j + 1] - sum[i];
    }
}