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42. 接雨水
接雨水这道题目是 面试中特别高频的一道题,也是单调栈 应用的题目,大家好好做做。
建议是掌握 双指针 和单调栈,因为在面试中 写出单调栈可能 有点难度,但双指针思路更直接一些。
在时间紧张的情况有,能写出双指针法也是不错的,然后可以和面试官在慢慢讨论如何优化。
class Solution { public int trap(int[] height) { int total = 0; for (int index = 0; index < height.length; index++) { if (index == 0 || index == height.length - 1) continue; int rightMax = height[index]; int leftMax = height[index]; for (int right = index + 1; right < height.length; right++) { rightMax = Math.max(rightMax, height[right]); } for (int left = index - 1; left >= 0; left--) { leftMax = Math.max(leftMax, height[left]); } int waterHeight = Math.min(leftMax, rightMax) - height[index]; if (waterHeight > 0) total += waterHeight; } return total; } }
- 柱状图中最大的矩形
有了之前单调栈的铺垫,这道题目就不难了。
class Solution { public int largestRectangleArea(int[] heights) { int[] extendedHeights = new int[heights.length + 2]; System.arraycopy(heights, 0, extendedHeights, 1, heights.length); extendedHeights[0] = 0; extendedHeights[heights.length + 1] = 0; Stack<Integer> stack = new Stack<>(); stack.push(0); int maxArea = 0; for (int i = 1; i < extendedHeights.length; i++) { while (!stack.isEmpty() && extendedHeights[i] < extendedHeights[stack.peek()]) { int topIndex = stack.pop(); int width = i - stack.isEmpty() ? 0 : i - stack.peek() - 1; int height = extendedHeights[topIndex]; maxArea = Math.max(maxArea, width * height); } stack.push(i); } return maxArea; } }