問題
Bars of width 1 have the given heights. After rain, water collects above each bar up to the lower of the tallest bars on its left and right. Compute the total trapped water. An O(n) two-pointer solution is expected.
Implement trapped_water(heights).
Input
heights: Bar heights from left to right.
Output
Return the total units of water.
制約
0 ≤ length ≤ 200,000; 0 ≤ height ≤ 100,000
例
例 1
入力:
heights = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
期待される出力:
expected = 6
入力:
std::vector<long long> heights = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
期待される出力:
long long expected = 6;
入力:
long[] heights = new long[] {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
期待される出力:
long expected = 6;
解説
Water fills the dips: 1 + 1 + 2 + 1 + 1 = 6 units.
例 2
入力:
heights = [3, 2, 1]
期待される出力:
expected = 0
入力:
std::vector<long long> heights = {3, 2, 1};
期待される出力:
long long expected = 0;
入力:
long[] heights = new long[] {3, 2, 1};
期待される出力:
long expected = 0;
解説
Heights only go down, so water runs off: 0.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。