Trapping Rain Water

问题描述

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given

[0,1,0,2,1,0,1,3,2,1,2,1]

, return

6

.



The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

解决思路

假设数组的长度为len，(1) 找到最高的那个柱子highest；

(2) 双指针：从0到highest, 从len-1到highest；

(3) 辅助栈s：存的是柱子高度的升序序列，如果遇到比栈顶元素小的高度，则能够储存水量为s.peek()-cur

时间空间复杂度均为O(n).

程序

public class Solution {
public int trap(int[] height) {
if (height == null || height.length == 0) {
return 0;
}
int highest = getHighestIdx(height);
int water = 0;

Stack<Integer> s = new Stack<Integer>();
for (int i = 0; i < highest; i++) {
if (s.isEmpty() || height[i] > s.peek()) {
s.push(height[i]);
} else {
water += s.peek() - height[i];
}
}

s = new Stack<Integer>();
for (int i = height.length - 1; i > highest; i--) {
if (s.isEmpty() || height[i] > s.peek()) {
s.push(height[i]);
} else {
water += s.peek() - height[i];
}
}

return water;
}

private int getHighestIdx(int[] height) {
int high = 0;
int idx = 0;
for (int i = 0; i < height.length; i++) {
if (height[i] > high) {
high = height[i];
idx = i;
}
}
return idx;
}
}