Rain Terraces / Trapping Rain Water
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
先从这个心智模型开始
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
它如何工作，不堆砌不必要的术语
A two-pointer solution tracks leftMax and rightMax while moving the side with the smaller boundary.
This gives O(n) time and O(1) extra space.
把算法读成一连串决策
Restate constraints and edge cases.
Identify the pattern and invariant.
Walk through a small example before coding.
Read a tiny input by hand and write down the state after every important operation. For Rain Terraces / Trapping Rain Water, the goal is to make the invariant visible rather than to memorize code.
function trap(height) {
let l = 0, r = height.length - 1;
let leftMax = 0, rightMax = 0, water = 0;
while (l < r) {
if (height[l] <= height[r]) {
leftMax = Math.max(leftMax, height[l]);
water += leftMax - height[l++];
} else {
rightMax = Math.max(rightMax, height[r]);
water += rightMax - height[r--];
}
}
return water;
}不要死记语法。跟踪每一行中发生变化的状态，并把它与上面描述的不变量联系起来。
Varies by implementation / input constraints
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
适用场景
- Use the recognized pattern when its invariant matches the problem constraints.
通常哪里会出错
- Jumping to code before defining the invariant.
- Optimizing before producing a correct baseline.
专业回答结构
Explain why moving the smaller boundary is safe: the smaller side already determines the maximum water level for that position.
What problem does Rain Terraces / Trapping Rain Water solve?
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
关于复杂度应该提到什么？
Its typical cost here is Varies by implementation / input constraints, but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
最常见的错误是什么？
Jumping to code before defining the invariant.