标签:面试 leetcode 收集雨水 trapping rain
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!
左边右边中短者为堤坝。比堤坝更短者可蓄一定的水。逐步收集这些水量。比当前堤坝更高者,将成为新堤坝。
此算法在leetcode上实际执行时间为10ms。
class Solution {
public:
int trap(int A[], int n) {
int low = 0, high = n-1;
int rain = 0, bar = 0;
while (low < high) {
if (A[low] < A[high]) {
if (bar < A[low])
bar = A[low];
else
rain += bar-A[low];
++low;
}
else {
if (bar < A[high])
bar = A[high];
else
rain += bar-A[high];
--high;
}
}
return rain;
}
};方法二,假定都是雨水,再逐步刨出那些堤坝占据的水量。
此算法在leetcode 上实际执行时间为13ms。比上面慢点,可能是因为用到了乘法。
class Solution {
public:
int trap(int A[], int n) {
int low = 0, high = n-1;
int bar = 0, rain = 0;
while (low < high) {
const int newbar = min(A[low], A[high]);
if (newbar > bar) {
rain += (newbar-bar) * (high-low);
bar = newbar;
}
if (A[low] < A[high]) {
rain -= min(bar, A[low]);
++low;
}
else {
rain -= min(bar, A[high]);
--high;
}
}
return rain;
}
};
Trapping Rain Water -- leetcode
标签:面试 leetcode 收集雨水 trapping rain
原文地址:http://blog.csdn.net/elton_xiao/article/details/42836505