Trapping Rain Water -- leetcode

標籤：面試   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!

方法一：收集雨水

左邊右邊中短者為堤壩。比堤壩更短者可蓄一定的水。逐步收集這些水量。比當前堤壩更高者，將成為新堤壩。

此演算法在leetcode上實際執行時間為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，再逐步刨出那些堤壩佔據的水量。

此演算法在leetcode 上實際執行時間為13ms。比上面慢點，可能是因為用到了乘法。

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