Leetcode42:trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar are 1, compute how much WA ter It is the 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 the Rain Water (blue section) is being trapped. Thanks Marcos for contributing this image!

The problem had been done a long time ago, and now it came to me, and soon it was solved.

Solution One

Two-time traversal method.

A traversal is performed first, the index of the highest element is found, then a second traversal is performed, and the second traversal is divided into a traverse from the left to the maximum and a right-to-maximum value.

When traversing from the left, a variable is used to record the maximal value starting from the left, and if the currently traversed element is less than this maximum, the difference before them is the amount of water that the current point can hold, and if the current element is greater than this maximum value, the maximum value is updated. The right side of the traversal method is similar. Such as:

Runtime:8ms

Class Solution {public:    int Trap (vector<int>& height) {        int maxpos=0;        int maxheight=0;        int length=height.size ();        for (int i=0;i<length;i++)        {            if (height[i]>maxheight)            {                maxheight=height[i];                maxpos=i;            }        }        int result=0;        int leftheight=0;        for (int i=0;i<maxpos;i++)        {            if (height[i]>leftheight)            {                leftheight=height[i];            }            else            {                result+=leftheight-height[i];            }        }        int rightheight=0;        for (int i=length-1;i>maxpos;i--)        {            if (height[i]>rightheight)            {                rightheight=height[i];            }            Else            {                result+=rightheight-height[i];            }        }        return result;}    ;

Solution Two

The second solution is to use a single traversal method.

The above solution is to traverse two times, in order to find the maximum value, so as to ensure that the difference between the maximal value and the current value is the amount of water can be accommodated, but in fact can not find the maximum value but the left and right of the current value of the maximum value. If the current value on the left is large, then traverse to the right, and if the current value on the right is large, walk to the left, so that the difference between the maximum and the current value is the amount of water that can be accommodated.

Runtime:8ms

     int Trap (vector<int>& height) {        int length=height.size ();        int result=0;        int leftpos=0;        int rightpos=length-1;        int leftmax=0;        int rightmax=0;        while (Leftpos<rightpos)        {            if (Height[leftpos]


 
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Leetcode42:trapping Rain Water