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.

Trapping Rain Water Example Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Solution: Keep two pointers, one from start, one from end, and use another two pointers to track the highest block in the left and in the right.

public int trapRainWater(int[] heights) {
    if(heights == null || heights.length == 0) {
        return 0;
    }
    int start = 0, end = heights.length - 1;
    int leftBar = heights[0], rightBar = heights[heights.length - 1];
    int res = 0;
    while(start < end) {
        if(heights[start] <= heights[end]) {
            if(heights[start] <= leftBar) {
                res += leftBar - heights[start];
                start++;
            }
            else {
                leftBar = heights[start];
            }
        }
        else {
            if(heights[end] <= rightBar) {
                res += rightBar - heights[end];
                end--;
            }
            else {
                rightBar = heights[end];
            }
        }
    }
    return res;
}