Leetcode-Java （四）

最近两个周在忙项目，所以一直没有刷题。今天接着来，走起。。。。

Plus One

Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.You may assume the integer do not contain any leading zero, except the number 0 itself.The digits are stored such that the most significant digit is at the head of the list.    public static int[] plusOne(int[] digits) {        int[] digits_new = new int[digits.length + 1];        int c = 1;        int i = digits.length - 1;        for (; i >= 0; i--) {            if ((digits[i] + c) > 9) {                digits_new[i + 1] = 0;                digits[i] = 0;            } else {                digits_new[i + 1] = c + digits[i];                digits[i] = c + digits[i];                c = 0;            }        }        if (c == 1) {            digits_new[0] = 1;            return digits_new;        } else {            return digits;        }    }    public static void plusOneTest() {        int[] digits = {0};        int[] digitis_new = plusOne(digits);        System.out.println(Arrays.toString(digitis_new));    }

Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?Note: Given n will be a positive integer.    public static int climbStairs(int n) {        if (n == 0) {            return 0;        }        if (n == 1) {            return 1;        }        int step_count = 1; //n个1        int two_max_count = n / 2;        for (int i = 1; i <= two_max_count; i++) {            step_count += calcSteps(n - i, i);        }        return step_count;    }    //解法二    public static int climbStairs1(int n) {        if (n == 1 || n == 2) {            return n;        }        return climbStairs1(n - 1) + climbStairs1(n - 2);    }    public static int calcSteps(int m, int n) {        BigInteger steps = BigInteger.valueOf(m);        BigInteger j = BigInteger.valueOf(n);        for (int i = 1; i < n; i++) {            steps = steps.multiply(BigInteger.valueOf(--m));            j = j.multiply(BigInteger.valueOf(n - i));        }        steps = steps.divide(j);        return steps.intValue();    }    public static void climbStairsTest() {        int n = 44;        int steps = climbStairs(n);        System.out.println("steps : " + steps);        steps = climbStairs1(n);        System.out.println("steps : " + steps);    }

Gray Code

The gray code is a binary numeral system where two successive values differ in only one bit.Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:00 - 001 - 111 - 310 - 2Note:For a given n, a gray code sequence is not uniquely defined.For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.格雷码数学公式： 整数n的格雷码为 n^(n/2)    public static List<Integer> grayCode(int n) {        List<Integer> grayCode = new ArrayList<Integer>();        int value = 1;        for (int i = 0; i < n; i++) {            value *= 2;        }        for (int i = 0; i < value; i++) {            grayCode.add(i ^ (i / 2));        }        return grayCode;    }    public static void grayCodeTest() {        int n = 2;        List<Integer> list = grayCode(n);        System.out.println(list);    }

Set Matrix Zeroes

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.click to show follow up.Follow up:Did you use extra space?A straight forward solution using O(mn) space is probably a bad idea.A simple improvement uses O(m + n) space, but still not the best solution.Could you devise a constant space solution?    public static void setZeroes(int[][] matrix) {        int[] row = new int[matrix.length];        int[] col = new int[matrix[0].length];        for (int i = 0; i < matrix.length; i++) {            for (int j = 0; j < matrix[0].length; j++) {                if (matrix[i][j] == 0) {                    row[i] = 1;                    col[j] = 1;                }            }        }        //清除行        for (int i = 0; i < matrix.length; i++) {            if (row[i] == 1) {                for (int j = 0; j < matrix[0].length; j++) {                    matrix[i][j] = 0;                }            }        }        //清除列        for (int j = 0; j < matrix[0].length; j++) {            if (col[j] == 1) {                for (int i = 0; i < matrix.length; i++) {                    matrix[i][j] = 0;                }            }        }    }    public static void setZeroesTest() {        int[][] matrix = {                {0, 0, 5, 8, 3, 8},                {9, 4, 1, 9, 9, 5},                {0, 4, 3, 0, 2, 7},                {1, 6, 0, 0, 3, 0},                {4, 4, 0, 3, 3, 7},                {0, 3, 7, 5, 1, 0}};        setZeroes(matrix);        for (int i = 0; i < matrix.length; i++) {            for (int j = 0; j < matrix[0].length; j++) {                System.out.print(matrix[i][j]);            }            System.out.println();        }    }

Gas Station

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.Note:The solution is guaranteed to be unique.public static int canCompleteCircuit(int[] gas, int[] cost) {        int total=0,tank=0,index=0;        for (int i = 0;i<gas.length;i++) {            tank += gas[i] - cost[i];            if (tank < 0){                index = i+1;                tank =0;            }            total +=gas[i] - cost[i];        }        return (total<0)?-1:index;    }    //这种解法时间发杂都有点大    //public static int canCompleteCircuit(int[] gas, int[] cost) {    //    int remainingPetrol = 0;    //    int N = gas.length;    //    int startPos = -1;    //    for (int i = 0; i < N; i++) {    //        startPos = i;    //        for (int m = i; m < N; m++) {    //            remainingPetrol += gas[m] - cost[m];    //            if (remainingPetrol < 0) {    //                startPos = -1;    //                break;    //            }    //        }    //        for (int n = 0; n < i; n++) {    //            remainingPetrol += gas[n] - cost[n];    //            if (remainingPetrol < 0) {    //                startPos = -1;    //                break;    //            }    //        }    //        if (startPos != -1) {    //            break;    //        }    //        remainingPetrol=0;    //    }    //    return startPos;    //}    public static void canCompleteCircuitTest(){        int[] gas = {2,3,1};        int[] cost = {3,1,2};        int startPos = canCompleteCircuit(gas,cost);        System.out.println("开始位置:"+startPos);    }