LeetCode 42 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!
根据输入的数组组成的直方图计算可以“盛水”的面积。idea是从第一个非零数开始标记为lastMarkedheight,这个lastMarkedheight作为一个可以盛水的高度基准,为了确定它是安全的,需要遍历它之后的数,要是有更高的数字则是安全的,否则把lastMarkedheight的值降低为可搜索到的最高高度。在确认安全之后继续遍历数组,遇到矮的则把高度差作为储水量加起来,遇到比lastMarkedheight高的就重新判断那个值是否安全并更新lastMarkedheight。
1 class Solution { 2 public: 3 int trap(vector<int>& height) { 4 vector<int>::iterator it; 5 int sumWater=0; 6 int lastMarkedHight=0; 7 for(it = height.begin();it!=height.end();it++){ 8 if(lastMarkedHight==0 && *it == 0)//前面的0无视掉 9 continue; 10 if(lastMarkedHight==0 && *it != 0){//第一个非零作为lastMarkedHight 11 lastMarkedHight = *it; 12 } 13 if(lastMarkedHight!=0 && *it>=lastMarkedHight){//判断是否安全 14 int a = 0; 15 bool findLarger = false; 16 for(vector<int>::iterator tempIt=it+1;tempIt!=height.end();tempIt++){//找是否有更高的 17 if(*tempIt>*it){ 18 findLarger = true; 19 break; 20 } 21 } 22 if(findLarger){//安全 23 lastMarkedHight = *it; 24 continue; 25 } 26 else{//找it后最高的一个数作为lastMarkedHight 27 while(find(it+1,height.end(),*it-a)==height.end() && *it-a>0){ 28 a++; 29 } 30 lastMarkedHight = *it-a; 31 } 32 continue; 33 } 34 if(lastMarkedHight!=0 && *it<lastMarkedHight && it!=height.end()-1){ 35 sumWater = sumWater+(lastMarkedHight-*it);//遇到矮的加储水量 36 cout<<sumWater<<endl; 37 continue; 38 } 39 } 40 return sumWater; 41 } 42 };