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