Trapping rain water

最新推荐文章于 2024-09-17 16:23:14 发布

转载最新推荐文章于 2024-09-17 16:23:14 发布 · 309 阅读

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#array #sum

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本文介绍了一种计算在下雨后，给定高度数组所形成的积水量的方法。通过找到最高柱子并分段计算每个区域的积水量，实现算法的高效运行。

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array的找规律的题

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.

public class Solution {
    public int trap(int[] A) {
      int sum = 0;
      int maxInd = -1;
      int max = -1;
      
      //find the highest bars;
      for (int i = 0; i < A.length; i++) {
          if (A[i] > max) {
              max = A[i];
              maxInd = i;
          }
      }
      
      //loop through the left part from 0 to i-1
      int pre = 0;
      for (int i = 0; i < maxInd; i++) {
          if (A[i] > pre){
              sum += (A[i] - pre) * (maxInd - i);
              pre = A[i];
          }
          sum -= A[i];
      }
      
      //loop through the right part from A.length to i+1
      pre = 0;
      for (int i = A.length-1; i > maxInd; i--) {
          if (A[i] > pre) {
              sum += (A[i] - pre) * (i - maxInd);
              pre = A[i];
          }
          sum -= A[i];
      }
      
      return sum;
    }
}