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LeetCode42 - 接雨水(难度:困难)
链接:https://leetcode-cn.com/problems/trapping-rain-water
给定 n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水。
示例 1:
输入:height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出:6
解释:上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图,在这种情况下,可以接 6 个单位的雨水(蓝色部分表示雨水)。
示例 2:
输入:height = [4,2,0,3,2,5]
输出:9
class Solution {
public int trap(int[] height) {
if(height == null || height.length == 0) return 0;
Stack<Integer> helpStack = new Stack<>();
int n = height.length;
// 左右边界
int[] left = new int[n];
int[] right = new int[n];
// Arrays.fill(right, n);
int maxHeight = -1;
// 遍历确定左右边界
for(int i=0;i<height.length; i++) {
while(!helpStack.isEmpty() && height[helpStack.peek()] < height[i]) {
helpStack.pop();
}
left[i] = helpStack.isEmpty()?-1:helpStack.peek();
if(helpStack.isEmpty() || height[helpStack.peek()] < height[i]){
helpStack.add(i);
}
if(maxHeight < height[i]) {
maxHeight = height[i];
}
}
helpStack.clear();
for(int j=height.length-1;j>=0; j--) {
while(!helpStack.isEmpty() && height[helpStack.peek()] <= height[j]) {
helpStack.pop();
}
right[j] = helpStack.isEmpty()?n:helpStack.peek();
if(helpStack.isEmpty() || height[helpStack.peek()] < height[j]){
helpStack.add(j);
}
}
int result = 0;
for(int i=0; i<height.length; i++) {
if(left[i] == -1 || right[i] == n) {
continue;
}
int abs = Math.abs(height[left[i]] - height[right[i]]);
result += (maxHeight - abs - height[i]);
}
return result;
}
}
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