多方法 05

42. 接雨水

https://leetcode-cn.com/problems/trapping-rain-water/

官方的题解很不错

https://leetcode-cn.com/problems/trapping-rain-water/solution/jie-yu-shui-by-leetcode-solution-tuvc/

动态规划

// dynamic programming
// Time O(n) Space O(n)
public int trap(int[] height) {
    int n = height.length;
    if (n == 0) return 0;
    int[] leftMax = new int[n];
    leftMax[0] = height[0];
    for (int i = 1; i < n; i++) {
        leftMax[i] = Math.max(height[i], leftMax[i - 1]);
    }
    int[] rigthMax = new int[n];
    rigthMax[n - 1] = height[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        rigthMax[i] = Math.max(height[i], rigthMax[i + 1]);
    }
    int res = 0;
    for (int i = 0; i < n; i++) {
        res += Math.min(leftMax[i], rigthMax[i]) - height[i];
    }
    return res;
}

单调栈

public int trap(int[] height) {
    int n = height.length;
    if (n == 0) return 0;
    Deque<Integer> stack = new LinkedList<>();
    int res = 0;
    for (int i = 0; i < n; i++) {
        while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
            int top = stack.pop();
            if (stack.isEmpty())
                break;
            int left = stack.peek();
            int curWidth = i - left - 1;
            int curHeight = Math.min(height[left], height[i]) - height[top];
            res += curHeight * curWidth;
        }
        stack.push(i);
    }
    return res;
}

双指针

// Double pointer
// Time O(n) Space O(1)
public int trap(int[] height) {
    int n = height.length;
    if (n == 0) return 0;
    int res = 0;
    int leftMax = 0, rightMax = 0, left = 0, right = n - 1;
    while (left < right) {
        leftMax = Math.max(leftMax, height[left]);
        rightMax = Math.max(rightMax, height[right]);
        if (height[left] < height[right]) {
            res += leftMax - height[left];
            left++;
        } else {
            res += rightMax - height[right];
            right--;
        }
    }
    return res;
}