Trapping Rain Water--LeetCode
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题目:
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!
思路:可以在这个序列中找到最高柱子的位置,那么从两头开始找可以盛水的多少,假如从前开始遍历,需要遍历到柱子最高的位置,遍历到当前位置,如果发现当前的柱子比之前记录的最高的柱子高,那么更新,如果没有之前记录的柱子高,那么就可以计算出当前柱子相对之前的高柱子盛水量。
方法二:也可以借助辅助栈,栈中记录的是位置,栈底始终是遍历到当前位置之前最高的柱子,如果发现当前比栈底柱子的高度还要打,那么需要计算从当前柱子到栈底这段空间可以盛水量,然后将当前为主压入栈,知道最后,栈空间可能不为空,这个时候需要清空栈,需要看当前栈中的柱子可以盛水量。
#include <iostream>#include <vector>#include <string> using namespace std;/*这个问题可以使用栈来 也可以不用 */int TrapRainWater(vector<int>& vec){int i,maxhigh;maxhigh = 0;int left=0,right = 0;int sum =0;for(i=0;i<vec.size();i++)if(vec[i] > vec[maxhigh])maxhigh = i;for(i=0;i<maxhigh;i++){if(vec[i] < left)sum +=(left-vec[i]);elseleft = vec[i];}for(i=vec.size()-1;i>maxhigh;i--){if(vec[i]<right)sum += (right-vec[i]);elseright = vec[i];}return sum;}int main(){int array[]={0,1,0,2,1,0,1,3,2,1,2,1};vector<int> vec(array,array+sizeof(array)/sizeof(int));cout<<TrapRainWater(vec)<<endl;}
两次遍历数组,第一次遍历找每个元素右边最大的元素,第二次遍历寻找每个元素左边最大的元素,同时计算该index可以存放的水容积: min{lefthighest, righthighest}-A[i]
public class Solution { public int trap(int[] A) { int[] left = new int[A.length]; int[] right = new int[A.length]; int sum = 0; for (int i=0; i<A.length; i++) { if (i == 0) left[i] = 0; else { left[i] = Math.max(left[i-1], A[i-1]); } } for (int i=A.length-1; i>=0; i--) { if (i == A.length - 1) right[i] = 0; else { right[i] = Math.max(right[i+1], A[i+1]); } if (Math.min(left[i], right[i]) > A[i]) { sum += Math.min(left[i], right[i]) - A[i]; } } return sum; }}
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