42. Trapping Rain Water

Problem

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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

Solution (Java)

class Solution {
    public int trap(int[] height) {
        int l = 0;
        int r = height.length - 1;
        int res = 0;
        int lowerWall = 0;
        while (l < r) {
            int lVal = height[l];
            int rVal = height[r];
            // If left is smaller than right ptr, make the lower wall the bigger of lVal and its
            // current size
            if (lVal < rVal) {
                // If lVal has gone up, move the lowerWall upp
                lowerWall = Math.max(lVal, lowerWall);
                // Add the water level at current point
                // Calculate this by taking the current value and subtracting it from the lower wall
                // size
                // We know that this is the lower wall because we've already determined that lVal <
                // rVal
                res += lowerWall - lVal;
                // Move left ptr along
                l++;
            } else {
                // Do the same thing, except now we know that the lowerWall is the right side.
                lowerWall = Math.max(rVal, lowerWall);
                res += lowerWall - rVal;
                r--;
            }
        }
        return res;
    }
}

Solution (Javascript)

var trap = function(height) {
  var res = 0;
  var left = 0;
  var right = height.length - 1;
  var leftMax = 0;
  var rightMax = 0;

  while (left < right) {
    if (height[left] < height[right]) {
      if (height[left] >= leftMax) {
        leftMax = height[left];
      } else {
        res += leftMax - height[left];
      }
      left++;
    } else {
      if (height[right] >= rightMax) {
        rightMax = height[right];
      } else {
        res += rightMax - height[right];
      }
      right--;
    }
  }

  return res;
};

Explain:

双指针

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).