牛客网左神算法中级班学习笔记（第三章）

本文是牛客网左神算法中级班学习笔记。

【分析】
宏观考虑！！！搞两个点A，B，起始都在左上角，B往右走，走到最右边就往下走，A往下走，走到最下边就往右走。A，B每次一起走一步，打印A，B两点连线即可（用一个Boolean控制下，交替打印顺序）。

public static void printMatrixZigZag(int[][] matrix) {
		int tR = 0;
		int tC = 0;
		int dR = 0;
		int dC = 0;
		int endR = matrix.length - 1;
		int endC = matrix[0].length - 1;
		boolean fromUp = false;
		while (tR != endR + 1) {
			printLevel(matrix, tR, tC, dR, dC, fromUp);
			tR = tC == endC ? tR + 1 : tR;
			tC = tC == endC ? tC : tC + 1;
			dC = dR == endR ? dC + 1 : dC;
			dR = dR == endR ? dR : dR + 1;
			fromUp = !fromUp;
		}
		System.out.println();
	}

	public static void printLevel(int[][] m, int tR, int tC, int dR, int dC,
			boolean f) {
		if (f) {
			while (tR != dR + 1) {
				System.out.print(m[tR++][tC--] + " ");
			}
		} else {
			while (dR != tR - 1) {
				System.out.print(m[dR--][dC++] + " ");
			}
		}
	}

【分析】
宏观考虑！！！搞两个点A，B，分别在左上角和右下角，打印外围一个框之后，就往里缩，一直打印即可。

public static void spiralOrderPrint(int[][] matrix) {
		int tR = 0;
		int tC = 0;
		int dR = matrix.length - 1;
		int dC = matrix[0].length - 1;
		while (tR <= dR && tC <= dC) {
			printEdge(matrix, tR++, tC++, dR--, dC--);
		}
	}

	public static void printEdge(int[][] m, int tR, int tC, int dR, int dC) {
		if (tR == dR) {
			for (int i = tC; i <= dC; i++) {
				System.out.print(m[tR][i] + " ");
			}
		} else if (tC == dC) {
			for (int i = tR; i <= dR; i++) {
				System.out.print(m[i][tC] + " ");
			}
		} else {
			int curC = tC;
			int curR = tR;
			while (curC != dC) {
				System.out.print(m[tR][curC] + " ");
				curC++;
			}
			while (curR != dR) {
				System.out.print(m[curR][dC] + " ");
				curR++;
			}
			while (curC != tC) {
				System.out.print(m[dR][curC] + " ");
				curC--;
			}
			while (curR != tR) {
				System.out.print(m[curR][tC] + " ");
				curR--;
			}
		}
	}

【分析】
依旧是宏观考虑！！！准备两个点A，B，起始位置分别是左上角和右下角。干完最外面的这个框框，然后往里面缩，依次干里面的框框即可。

public static void rotate(int[][] matrix) {
		int tR = 0;
		int tC = 0;
		int dR = matrix.length - 1;
		int dC = matrix[0].length - 1;
		while (tR < dR) {
			rotateEdge(matrix, tR++, tC++, dR--, dC--);
		}
	}

	public static void rotateEdge(int[][] m, int tR, int tC, int dR, int dC) {
		int times = dC - tC; 
		int tmp = 0;
		for (int i = 0; i != times; i++) {
			tmp = m[tR][tC + i];
			m[tR][tC + i] = m[dR - i][tC];
			m[dR - i][tC] = m[dR][dC - i];
			m[dR][dC - i] = m[tR + i][dC];
			m[tR + i][dC] = tmp;
		}
	}

【分析】
这个题目是剑指offer的！！！可以从右上角开始，如果小于给定值就往左找，如果大于给定值就往下找，直到找到为止，如果越界还没找到就说明不存在了！时间复杂度从O(M*N)优化到了O(M+N)。

public static boolean isContains(int[][] matrix, int K) {
		int row = 0;
		int col = matrix[0].length - 1;
		while (row < matrix.length && col > -1) {
			if (matrix[row][col] == K) {
				return true;
			} else if (matrix[row][col] > K) {
				col--;
			} else {
				row++;
			}
		}
		return false;
	}