LeetCode 282. Expression Add Operators(表达式)

原题网址:https://leetcode.com/problems/expression-add-operators/

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +-, or * between the digits so they evaluate to the target value.

Examples: 

"123", 6 -> ["1+2+3", "1*2*3"] 
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []
方法:深度优先搜索。

public class Solution {
    private void find(char[] na, int step, Expression[] exprs, int[] operands, int pos, int target, List results) {
        if (step == na.length) {
            if (exprs[pos-1] != Expression.OPERAND) return;
            // evaluate expressions
            int[] tempo = new int[5];
            Expression[] tempe = new Expression[5];
            int num = 0;
            int size = 0;
            for(int i=0; i0 && tempe[size-1]==Expression.MULTIPLY) {
                    tempo[size-2] *= operands[i];
                    size --;
                } else if (size>3 && (tempe[3]==Expression.PLUS || tempe[3]==Expression.MINUS)) {
                    if (tempe[1]==Expression.PLUS) {
                        tempo[0] += tempo[2];
                        tempe[1]=tempe[3];
                        tempo[2]=operands[i];
                        size=3;
                    } else {
                        tempo[0]-=tempo[2];
                        tempe[1]=tempe[3];
                        tempo[2]=operands[i];
                        size=3;
                    }
                } else {
                    tempo[size] = operands[i];
                    tempe[size] = exprs[i];
                    size ++;
                }
            }
            if (size==3) {
                if (tempe[1]==Expression.PLUS) {
                    tempo[0] += tempo[2];
                } else {
                    tempo[0]-=tempo[2];
                }
            }
            if (tempo[0]==target) {
                StringBuilder sb = new StringBuilder();
                for(int i=0; iInteger.MAX_VALUE) break;
                operands[pos] = (int)num;
                find(na, i+1, exprs, operands, pos+1, target, results);
                if (na[step]=='0') break;
            }
        } else {
            exprs[pos] = Expression.PLUS;
            find(na, step, exprs, operands, pos+1, target, results);
            exprs[pos] = Expression.MINUS;
            find(na, step, exprs, operands, pos+1, target, results);
            exprs[pos] = Expression.MULTIPLY;
            find(na, step, exprs, operands, pos+1, target, results);
        }
    }
    public List addOperators(String num, int target) {
        List results = new ArrayList<>();
        if (num == null || num.length()==0) return results;
        char[] na = num.toCharArray();
        find(na, 0, new Expression[na.length*2], new int[na.length*2], 0, target, results);
        return results;
    }
}
enum Expression {OPERAND, PLUS, MINUS, MULTIPLY};


你可能感兴趣的:(深度优先搜索,表达式运算,表达式,组合)