LeetCode 42. 接雨水

题目链接：42. 接雨水

class Solution {
    //暴力法
    public int trap(int[] height) {
        int ans = 0;
        int size = height.length;
        for (int i = 1; i < size - 1; i++) {
            int max_left = 0, max_right = 0;
            for (int j = i; j >= 0; j--) { //Search the left part for max bar size
                max_left = Math.max(max_left, height[j]);
            }
            for (int j = i; j < size; j++) { //Search the right part for max bar size
                max_right = Math.max(max_right, height[j]);
            }
            ans += Math.min(max_left, max_right) - height[i];
        }
        return ans;
    }

    //动态编程法
    public int trap1(int[] height) {
        if (height == null || height.length == 0)
            return 0;
        int ans = 0;
        int size = height.length;
        //leftMax[i]表示height[i]及其左边位置最高的位置
        int[] leftMax = new int[size];
        //rightMax[i]表示height[i]及其右边位置最高的位置
        int[] rightMax = new int[size];
        leftMax[0] = height[0];
        for (int i = 1; i < size; i++) {
            leftMax[i] = Math.max(height[i], leftMax[i - 1]);
        }
        rightMax[size - 1] = height[size - 1];
        for (int i = size - 2; i >= 0; i--) {
            rightMax[i] = Math.max(height[i], rightMax[i + 1]);
        }
        //leftMax[i]和rightMax[i]比较谁小，减去当前的高度就是水的容量
        for (int i = 1; i < size - 1; i++) {
            ans += Math.min(leftMax[i], rightMax[i]) - height[i];
        }
        return ans;
    }


    //双指针法
    public int trap2(int[] height) {
        int left = 0, right = height.length - 1;
        int ans = 0;
        int left_max = 0, right_max = 0;
        while (left < right) {
            if (height[left] < height[right]) {
                if (height[left] >= left_max) {
                    left_max = height[left];
                } else {
                    ans += (left_max - height[left]);
                }
                ++left;
            } else {
                if (height[right] >= right_max) {
                    right_max = height[right];
                } else {
                    ans += (right_max - height[right]);
                }
                --right;
            }
        }
        return ans;
    }
}