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:
- Thinking of using advanced data structures? You are thinking it too complicated.
- For each update operation, do you really need to update all elements between i and j?
- Update only the first and end element is sufficient.
- The optimal time complexity is O(k + n) and uses O(1) extra space.
分析¶
LintCode链接。暴力法:每次更新[startIndex, endIndex]区间
public int[] getModifiedArray(int length, int[][] updates) { int[] res = new int[length]; for (int i = 0; i < updates.length; i++) for (int j = updates[i][0]; j <= updates[i][1]; j++) res[j] += updates[i][2]; return res; }
比较巧妙的方法:
public int[] getModifiedArray(int length, int[][] updates) { int[] arr = new int[length + 1], res = new int[length]; for (int[] update : updates) { arr[update[0]] += update[2]; arr[update[1] + 1] -= update[2]; } res[0] = arr[0]; for (int i = 1; i < length; i++) res[i] = res[i - 1] + arr[i]; return res; }
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