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:

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

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树状数组，关于树状数组的详情可以参照http://baike.baidu.com/link?url=U7QI5Dr2xEufw9MuPWWTRQQuQCbdg9c5S6vXlDcweVUrT8z9RoQWC0qhjVQJUjPypWTOl4QGwe67QlQqz7u10q

public class NumArray {
int[] processed;
int[] nums;
int length;

public NumArray(int[] nums) {
length = nums.length;
processed = new int[length+1];
this.nums = nums;

//init processed
for(int i = 1;i<=length;i++){
int sum = 0;
int count = 1;
int counter = lowBit(i);

while(count <= counter){
sum += nums[i-count];
count++;
}
processed[i] = sum;
}
}

void update(int i, int val) {
//更新树状数组
int gap = val - nums[i];
nums[i] = val;

int index = i+1;
while(index <= length){
processed[index] += gap;
index += lowBit(index);
}
}

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

private int sum(int index){
int sum = 0;
while(index > 0){
sum += processed[index];
index -= lowBit(index);
}
return sum;
}
private int lowBit(int index){
return index & (-index);
}
}

// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);