跳到主要内容

Trapping Rain Water

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.

For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Trapping Rain Water

分析​

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是min(left_peak, right_peak) - height。

代码​

// Trapping Rain Water
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
  public: int trap(vector < int > & height) {
    int left = 0, right = height.size() - 1;
    int result = 0;
    int left_peak = 0, right_peak = 0;
    while (left < right) {
      if (height[left] < height[right]) {
        if (height[left] >= left_peak) {
          left_peak = height[left];
        } else {
          result += (left_peak - height[left]);
        }
        ++left;
      } else {
        if (height[right] >= right_peak) {
          right_peak = height[right];
        } else {
          result += (right_peak - height[right]);
        }
        --right;
      }
    }
    return result;
  }
};

相关题目​