【算法】【线性表】【数组】Trapping Rain Water(接水量)
1 题目
Given n
non-negative integers representing an elevation map where the width of each bar is 1
, compute how much water it can trap after raining.
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6 Explanation: The above elevation map (black section) 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.
Input: height = [4,2,0,3,2,5] Output: 9
Constraints:
n == height.length
1 <= n <= 2 * 104
0 <= height[i] <= 105
2 解答
2.1 比较好理解的
总的接水量 = 每个元素的接水量的和,而每个元素的接水量 = min(leftMax, rightMax) - 自己
class Solution { public int trap(int[] height) { // 记录总的接水量 int res = 0; // 总的接水量 = 每个元素的接水量的和 // 每个元素的接水量 = min(leftMax, rightMax) - 自己 // 求每个元素的左侧最大值 int len = height.length; int[] leftMax = new int[len]; leftMax[0] = height[0]; for (int i=1; i<len; i++) { leftMax[i] = Math.max(leftMax[i-1], height[i]); } // 求每个元素的右侧最大值 int[] rightMax = new int[len]; rightMax[len-1] = height[len-1]; for (int i=len-2; i>=0; i--) { rightMax[i] = Math.max(rightMax[i+1], height[i]); } // 计算每个元素的接水量 for (int i=0; i<len; i++) { res += Math.min(leftMax[i], rightMax[i]) - height[i]; } return res; } }