题目:
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 ]
解法:
这题与算法无关,是个数学题。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 i = 0; i < updates.length; i++) {
int start = updates[i][0], end = updates[i][1];
int inc = updates[i][2];
result[start] += inc;
if (end < length - 1) {
result[end + 1] -= inc;
}
}
int sum = 0;
for (int i = 0; i < length; i++) {
sum += result[i];
result[i] = sum;
}
return result;
}
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