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  • Trapping Rain Water

    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!

    C++实现代码:

    #include<iostream>
    using namespace std;
    
    class Solution {
    public:
         // 关键是在于找到规律:  
        // 即第i块地方的存水量 = min(第i块左边最高的bar高度, 第i块右边最高的bar的高度) - 第i块地方bar的高度  
        // 例如图中,第5块地方的存水量 = min(2,3)-0 = 2  
        // 2为其左边最高的bar,即第3块地方的bar  
        // 3为其右边最高的bar,即第7块地方的bar,  
        // 0为其自身的bar高度 
        int trap(int A[], int n) {
            if(n==0)
                return 0;
            int left[n];
            int right[n];
            int i;
            int sum=0;
            left[0]=A[0];
            for(i=1;i<n;i++)
                left[i]=max(left[i-1],A[i]);
            right[n-1]=A[n-1];
            for(i=n-2;i>=0;i--)
                right[i]=max(right[i+1],A[i]);
        //注意边界不用最后一个和第一个都不需要
    for(i=1;i<n-1;i++) sum+=(min(left[i],right[i])-A[i]); return sum; } }; int main() { Solution s; int A[12]={0,1,0,2,1,0,1,3,2,1,2,1}; cout<<s.trap(A,12)<<endl; }

    参考:http://blog.csdn.net/fightforyourdream/article/details/15026089

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  • 原文地址:https://www.cnblogs.com/wuchanming/p/4121496.html
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