Leetcode 04: T26. Delete duplicate items in the sorting array (simple); Sword finger offer 67. convert the string to an integer (medium); Interview question 01.08. zero matrix (simple)

T10: 26. Delete duplicates in sort array （ Simple ）

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas :

solution : Double pointer

class Solution {
    
    public int removeDuplicates(int[] nums) {
    
        int n = nums.length;
        if (n == 0) {
    
            return 0;
        }
        int fast = 1, slow = 1;
				// Why not slow = 0
        while (fast < n) {
    
            if (nums[fast] != nums[fast - 1]) {
    
                nums[slow] = nums[fast];
                ++slow;
            }
            ++fast;
        }
        return slow;
    }
}

Execution time ：1 ms, In all Java Defeated in submission **81.21%** Users of

Memory consumption ：40.1 MB, In all Java Defeated in submission **82.57%** Users of

T11: The finger of the sword Offer 67. Convert a string to an integer （ secondary ）

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/ba-zi-fu-chuan-zhuan-huan-cheng-zheng-shu-lcof
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas :

solution : I don't understand

class Solution {
    
    public int strToInt(String str) {
    
        char[] c = str.trim().toCharArray();
        if(c.length == 0) return 0;
        int res = 0, bndry = Integer.MAX_VALUE / 10;
        int i = 1, sign = 1;
        if(c[0] == '-') sign = -1;
        else if(c[0] != '+') i = 0;
        for(int j = i; j < c.length; j++) {
    
            if(c[j] < '0' || c[j] > '9') break;
            if(res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            res = res * 10 + (c[j] - '0');
        }
        return sign * res;
    }
}

Execution time ：2 ms, In all Java Defeated in submission **99.43%** Users of

Memory consumption ：38.5 MB, In all Java Defeated in submission **29.82%** Users of

T12: Interview questions 01.08. Zero matrix （ Simple ）

Write an algorithm , if M × N An element in the matrix is 0, The row and column in which it is located are cleared .

Example 1：

 Input ：
[
[1,1,1],
[1,0,1],
[1,1,1]
]
 Output ：
[
[1,0,1],
[0,0,0],
[1,0,1]
]

Example 2：

 Input ：
[
[0,1,2,0],
[3,4,5,2],
[1,3,1,5]
]
 Output ：
[
[0,0,0,0],
[0,4,5,0],
[0,3,1,0]
]

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/zero-matrix-lcci
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas

solution : boolean Mark

class Solution {
    
    public void setZeroes(int[][] matrix) {
    
        int m = matrix.length, n = matrix[0].length;
        boolean[] row = new boolean[m];
        boolean[] col = new boolean[n];
        for (int i = 0; i < m; i++) {
    
            for (int j = 0; j < n; j++) {
    
                if (matrix[i][j] == 0) {
    
                    row[i] = col[j] = true;
                }
            }
        }
        for (int i = 0; i < m; i++) {
    
            for (int j = 0; j < n; j++) {
    
                if (row[i] || col[j]) {
    
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

Execution time ：1 ms, In all Java Defeated in submission **99.84%** Users of

Memory consumption ：40.1 MB, In all Java Defeated in submission **52.76%** Users of

Time complexity ：O(mn), among m It's the number of rows in the matrix ,n It's the number of columns in a matrix . We only need to traverse the matrix twice at most .

Spatial complexity ：O(m+n), among m It's the number of rows in the matrix ,n It's the number of columns in a matrix . We need to record whether there is zero in each row or column .