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.
1 class Solution { 2 public: 3 int trap(int A[], int n) { 4 int sum = 0; 5 int *max_left = new int[n](); 6 for (int i = 1; i < n - 1; ++i) { 7 if (A[i - 1] > max_left[i - 1]) { 8 max_left[i] = A[i - 1]; 9 } else { 10 max_left[i] = max_left[i - 1]; 11 } 12 } 13 int max_right = 0; 14 for (int i = n - 2; i >= 1; --i) { 15 if (A[i + 1] > max_right) { 16 max_right = A[i + 1]; 17 } 18 if (A[i] < max_left[i] && A[i] < max_right) { 19 sum += (min(max_left[i], max_right) - A[i]); 20 } 21 } 22 return sum; 23 } 24 };
单独考虑每根bar，每根bar上方可以盛的水量取决于从它向左扫描到的最高的bar和向右扫描到的最高的bar中较矮的那一个和它本身的高度差。
扫描一遍数组，记录下每根bar左边最高的bar，然后从右向左，一遍记录当前最高的bar，一边计算每根bar上方可以盛水的量，并求和。