描述
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.
如上图所示,海拔分别为 [0,1,0,2,1,0,1,3,2,1,2,1], 返回 6.
挑战
O(n) 时间, O(1) 空间
O(n) 时间, O(n) 空间也可以接受
分析
找到最高的那个位置,然后分两边进行计算。
程序
class Solution { public: /** * @param heights: a list of integers * @return: a integer */ int trapRainWater(vector<int> &heights) { // write your code here //找到最高的那个分水岭 int maxheights, maxindex; maxheights = 0; for(int i = 0; i < heights.size(); i++){ if(maxheights < heights[i]){ maxindex = i; maxheights = heights[i]; } } //分水岭右边 int sum = 0; maxheights = 0; for(int i = 0; i < maxindex; i++){ if(maxheights > heights[i]) sum += maxheights - heights[i]; maxheights = max(maxheights, heights[i]); } //分水岭左边 maxheights = 0; for(int i = heights.size() - 1; i > maxindex; i--){ if(maxheights > heights[i]) sum += maxheights - heights[i]; maxheights = max(maxheights, heights[i]); } return sum; } };