Thursday, June 30, 2016

Leetcode 370 - Range Addition


http://www.cnblogs.com/grandyang/p/5628786.html
Assume you have an array of length n initialized with all 0's and are given k update operations.
Each operation is represented as a triplet: [startIndex, endIndex, inc] which increments each element of subarray A[startIndex ... endIndex] (startIndex and endIndex inclusive) with inc.
Return the modified array after all k operations were executed.
Example:
Given:

    length = 5,
    updates = [
        [1,  3,  2],
        [2,  4,  3],
        [0,  2, -2]
    ]

Output:

    [-2, 0, 3, 5, 3]
Explanation:
Initial state:
[ 0, 0, 0, 0, 0 ]

After applying operation [1, 3, 2]:
[ 0, 2, 2, 2, 0 ]

After applying operation [2, 4, 3]:
[ 0, 2, 5, 5, 3 ]

After applying operation [0, 2, -2]:
[-2, 0, 3, 5, 3 ]
Hint:
  1. Thinking of using advanced data structures? You are thinking it too complicated.
  2. For each update operation, do you really need to update all elements between i and j?
  3. Update only the first and end element is sufficient.
  4. The optimal time complexity is O(k + n) and uses O(1) extra space.
这道题的提示说了我们肯定不能把范围内的所有数字都更新,而是只更新开头结尾两个数字就行了,那么我们的做法就是在开头坐标startIndex位置加上inc,而在结束位置加1的地方加上-inc,那么根据题目中的例子,我们可以得到一个数组,nums = {-2, 2, 3, 2, -2, -3},然后我们发现对其做累加和就是我们要求的结果result = {-2, 0, 3, 5, 3}
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res, nums(length + 1, 0);
        for (int i = 0; i < updates.size(); ++i) {
            nums[updates[i][0]] += updates[i][2];
            nums[updates[i][1] + 1] -= updates[i][2];
        }
        int sum = 0;
        for (int i = 0; i < length; ++i) {
            sum += nums[i];
            res.push_back(sum);
        }
        return res;
    }
我们可以在空间上稍稍优化下上面的代码,用res来代替nums,最后把res中最后一个数字去掉即可
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res(length + 1);
        for (auto a : updates) {
            res[a[0]] += a[2];
            res[a[1] + 1] -= a[2];
        }
        for (int i = 1; i < res.size(); ++i) {
            res[i] += res[i - 1];
        }
        res.pop_back();
        return res;
    }
https://all4win78.wordpress.com/2016/06/29/leetcode-370-range-addition/
https://segmentfault.com/a/1190000005948569
思想是把所有需要相加的值存在第一个数,然后把这个范围的最后一位的下一位减去这个inc, 这样我所以这个范围在求最终值的时候,都可以加上这个inc,而后面的数就不会加上inc。

    public int[] getModifiedArray(int length, int[][] updates) {
        int[] result = new int[length];
        for (int[] operation : updates) {
            result[operation[0]] += operation[2];
            if (operation[1] < length - 1) {
                result[operation[1] + 1] -= operation[2];
            }
        }
        int temp = 0;
        for (int i = 0; i < length; i++) {
            temp += result[i];
            result[i] = temp;
        }
        return result;
    }
https://discuss.leetcode.com/topic/49674/java-o-n-k-time-o-1-space-with-algorithm-explained/
segment [i,j] is made of two parts [0,i-1] and [0, j]
so [i,j] increase 2 is same as [0,j] increase 2 and [0,i-1] increase -2. so you only need to update value at nums[j] with inc and nums[i-1] -inc. initially nums[i] is defined as all elements [0,i] increases inc
then think from length-1 to 0 backward. The last spot nums[length-1] does not need any modification.
nums[length-2] value should be updated as nums[length-2] + nums[length-1] as the latter covers the front. but front does not influence what is after it. so every spot should be updated as + the accumulate sum from the end.
    public int[] getModifiedArray(int length, int[][] updates) {
        int[] nums = new int[length];
        for (int[] update : updates) {
            nums[update[1]] += update[2];
            if (update[0] > 0) {
                nums[update[0] - 1] -= update[2];
            } 
        }
        
        int sum = nums[length - 1];
        for (int i = length - 2; i >= 0; i--) {
            int tmp = sum + nums[i];
            nums[i] += sum;
            sum = tmp; 
        }
        return nums;
    }
http://blog.csdn.net/jmspan/article/details/51787011
方法一:类似城市天际线问题,使用最小堆来维护当前的范围。
    public int[] getModifiedArray(int length, int[][] updates) {
        Arrays.sort(updates, new Comparator<int[]>() {
            @Override
            public int compare(int[] seg1, int[] seg2) {
                return Integer.compare(seg1[0], seg2[0]);
            }
        });
        PriorityQueue<Integer> heap = new PriorityQueue<Integer>(new Comparator<Integer>() {
            @Override
            public int compare(Integer i1, Integer i2) {
                return Integer.compare(updates[i1][1], updates[i2][1]);
            }
        });
        int[] results = new int[length];
        int j = 0;
        int sum = 0;
        for(int i = 0; i < length; i++) {
            while (!heap.isEmpty() && updates[heap.peek()][1] < i) {
                int p = heap.poll();
                sum -= updates[p][2];
            }
            while (j < updates.length && i >= updates[j][0]) {
                sum += updates[j][2];
                heap.offer(j);
                j++;
            }
            results[i] = sum;
        }
        return results;
    }
X. Brute force:
https://discuss.leetcode.com/topic/49669/java-straightforward-solution
public int[] getModifiedArray(int length, int[][] updates) {
    int[] nums = new int[length];
    int k = updates.length;
 for(int i = 0; i < k; i++){
  int start = updates[i][0];
  int end = updates[i][1];
  int inc = updates[i][2];
  for(int j = start; j <= end; j++){
   nums[j] += inc;
  }
 }
 return nums;
}

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