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【LeetCode】Trapping Rain Water
Trapping Rain Water
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. Thanks Marcos for contributing this image!
双指针i,j分别从首尾开始扫，记当前i指针遇到的最大值为leftWall，j指针遇到的最大值为rightWall
（1）leftWall <= rightWall
对于i指针指向的位置，被trap的值为(leftWall-A[i])。
i前进一个位置。
解释如下：
a.如果i与j之间不存在比leftWall大的值，那么i位置trap的值就取决与leftWall与rightWall的较小值，也就是leftWall
b.如果i与j之间存在比leftWall大的值，其中离leftWall最近的记为newLeftWall，那么i位置trap的值就取决与leftWall与newLeftWall的较小值，也就是leftWall
（2）leftWall > rightWall
对于j指针指向的位置，被trap的值为(rightWall-A[j])。
j前进一个位置。
解释同上。
class Solution {public: int trap(int A[], int n) { if(n < 3) //at least 3 integers return 0; int i = 0; int j = n-1; int leftWall = 0; int rightWall = 0; int ret = 0; while(i <= j) { //update leftWall and rightWall leftWall = max(leftWall, A[i]); rightWall = max(rightWall, A[j]); if(leftWall <= rightWall) {//no matter whether there is a more higher wall between leftWall and rightWall //A[i] is dependent on leftWall if leftWall <= rightWall ret += (leftWall-A[i]); i ++; } else {//no matter whether there is a more higher wall between leftWall and rightWall //A[j] is dependent on rightWall if rightWall < leftWall ret += (rightWall-A[j]); j --; } } return ret; }};
【LeetCode】Trapping Rain Water