leetcode 307. Range Sum Query - Mutable

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.

Example:

Given nums = [1, 3, 5]

sumRange(0, 2) -> 9
update(1, 2)
sumRange(0, 2) -> 8

Note:

1. The array is only modifiable by the update function.
2. You may assume the number of calls to update and sumRange function is distributed evenly.

---

======================2019.1.24==========================

节点版本，可以在每个节点维护区间和，还有区间最大值。

class NumArray {
    class SegmentTree{
        SegmentTree left;
        SegmentTree right;
        int sum;
    }
    int[] nums;
    SegmentTree root;
    public NumArray(int[] nums) {
        this.nums=nums;
        root=build(0,nums.length-1);
    }
    
    public SegmentTree build(int l,int r){
        if(l>r) return null;
        if(l==r){
            SegmentTree node=new SegmentTree();
            node.sum=nums[l];
            return node;
        }
        int m=(l+r)/2;
        SegmentTree node=new SegmentTree();
        node.left=build(l,m);
        node.right=build(m+1,r);
        node.sum=(node.left==null?0:node.left.sum)+(node.right==null?0:node.right.sum);
        return node;
    }
    
    public void update(int i, int val) {
        int dif=nums[i]-val;//需要减去
        nums[i]=val;
        update(root,i,0,nums.length-1,dif);
    }
    void update(SegmentTree node,int i,int l,int r,int dif){//root代表[0,nums.length-1]区间
        if(node==null) return;
        node.sum-=dif;
        if(l==r) return;
        int m=(l+r)/2;
        if(i>m){
            update(node.right,i,m+1,r,dif); 
        }else{
            update(node.left,i,l,m,dif);
        }
    }
    
    
    public int sumRange(int l, int r) {
        return sumRange(root,l,r,0,nums.length-1);
    }
    int sumRange(SegmentTree node,int l,int r,int ll,int rr){
        int mid=(ll+rr)/2;
        if(l==ll&&r==rr) return node.sum;
        int res=0;
        if(l<=mid&&r>mid) res+=sumRange(node.left,l,mid,ll,mid)+sumRange(node.right,mid+1,r,mid+1,rr);
        else if(r<=mid){
            res+=sumRange(node.left,l,r,ll,mid);    
        }else if(l>mid){
            res+=sumRange(node.right,l,r,mid+1,rr);
        }
        return res;
    }
}

 

========================================================

 

======================2019.1.23==========================

class NumArray {
    int[] nums;
    int[] sum;
    public NumArray(int[] nums) {
        this.nums=nums;
        sum=new int[nums.length*2];
        for(int i=nums.length;i<sum.length;i++){
            sum[i]=nums[i-nums.length];
        }
        for(int i=nums.length-1;i>0;i--) sum[i]=sum[i*2]+sum[i*2+1];
    }
    
    public void update(int i, int val) {
        int index=i+nums.length;
        int dif=nums[i]-val;
        nums[i]=val;
        while(index>0){
            sum[index]-=dif;
            index/=2;
        }
    }
    
    public int sumRange(int i, int j) {
        int l=i+nums.length,r=j+nums.length;
        int res=0;
        while(true){
            if(r%2==0){
                res+=sum[r];
                r--;
            }
            if(l%2==1){
                res+=sum[l];
                l++;
            }
            if(l>r) break;
            l=l/2;
            r=r/2;
        }
        return res;
    }
}

========================================================

参考https://blog.youkuaiyun.com/zearot/article/details/52280189 写了线段树，但看到leetcode参考答案还有更高效的写法（只需开辟2n数组长度，左右子节点区间不一定相同，自底向上迭代，能快速定位的叶子节点），待改进。

这里的线段树思路用sum[ ]保存各个区间和。sum一个下标对应一个区间。

class NumArray {
    int[] sum,nums;
    public NumArray(int[] nums) {
        if(nums==null||nums.length==0){
            this.nums=nums;
            return;
        }
        int len=nums.length;
        this.nums=nums;
        sum=new int[len*4];
        build(1,0,len-1);
    }
    
    //n为sum的下标，区间[l,r]对应n,构造线段树
    private void build(int n,int l,int r){
        if(l==r){
            sum[n]=nums[l];
            return;
        }
        int m=(l+r)/2;
        build(n*2,l,m);
        build(n*2+1,m+1,r);
        sum[n]=sum[2*n]+sum[2*n+1];
    }
        
    public void update(int i, int val) {
        update(i,val-nums[i],1,0,nums.length-1);
        nums[i]=val;
    }
    
    private void update(int index,int v,int n,int l,int r) {//l,r区间，index下标，v增加值，n区间对应的sum下标
        
        sum[n]+=v;
        if(l==r) return;
        int m=(l+r)/2;
        if(index>m){//在右边区间，区间对应index
            update(index,v,2*n+1,m+1,r);    
        }else{
            update(index,v,2*n,l,m); 
        }
    }
    
    public int sumRange(int i, int j) {
        return count(i,j,1,0,nums.length-1);
    }
    
    private int count(int i,int j,int index,int l,int r){
        if(i==l&&j==r){
            return sum[index];
        }
        
        int m=(l+r)/2;
        if(i>m){//在右边
            return count(i,j,index*2+1,m+1,r);
        }
        if(j<=m){//在左边
            return count(i,j,index*2,l,m);
        }
        //在两边
        return count(i,m,index*2,l,m)+count(m+1,j,index*2+1,m+1,r);
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * obj.update(i,val);
 * int param_2 = obj.sumRange(i,j);
 */

 

改进：参考https://leetcode.com/problems/range-sum-query-mutable/solution/

这样build线段树，树中的线段元素（比如共有5个元素的情况）可能有不连续的情况，但能节省空间。因为能定位到某个点，这是一种自底向上的方式，并没有上面显式的有一个sum的一个下标对应一个区间。

class NumArray {
    int[] nums,sum;
    int len;
    public NumArray(int[] nums) {
        this.nums=nums;
        len=nums.length;
        sum=new int[len<<1];
        //build树
        for(int i=len;i<len<<1;i++){
            sum[i]=nums[i-len];
        }
        for(int i=len-1;i>0;i--){
            sum[i]=sum[i<<1]+sum[i<<1|1];//左子节点偶数，右子节点奇数
        }
    }
    
    public void update(int i, int val) {
        int k=val-nums[i];
        nums[i]=val;
        int n=i+len;
        while(n>0){
            sum[n]+=k;//它的根节点都要更新
            n=n/2;
        }
    }
    
    public int sumRange(int i, int j) {
        int l=i+len,r=j+len;
        //左子节点偶数，右子节点奇数
        int s=0;
        while(l<=r){
            if(l%2==1){
                s+=sum[l];
                l++;
            } 
            if(r%2==0){
                s+=sum[r];
                r--;
            }    
            l=l/2;
            r=r/2;
        }
        return s;
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * obj.update(i,val);
 * int param_2 = obj.sumRange(i,j);
 */