[LeetCode], solution, non-code implementation

42. Trapping Rain Water

we need to find how many waters can each bar[i] trap. So we need to find the left peak from barto bar[i-1] and find the right peak from bar[i+1] to bar[n-1]. See the following pseudo-code:

total_water = 0; 

for i = 0 to n-1:

    Lmax = Max(h[0], h[1], ... h[i-1]);　　

    Rmax = Max(h[i+1], h[i+2], ... h[n-1]);

    water_and_building = Min(Lmax, Rmax);　　// water should not exceed the lowest height

    wi = water_and_building - h[i]; // the building has some height

    total_water += wi;

return total_water;